UC-NRLF $B 3Db 32^ OF THE UNIVERSITY OF CALIFOR^ EDDCATIOU IIBHj A & >♦» X THE Jjk I SITED STATES CALCULATOR; OR, ARITHMETIC SIMPLIFIED, IN DOLLARS AND CENTS, ADAPTED TO THE COMMERCE OF THE UNITED STATES, IN ITS FOREIGN AND DOMESTIC RELATIONS, ABBREVIATED, SIMPLIFIED, AND ARRANGED ON A NEW SY8TBM, IX A SERIES OF LECTURES, COMPRISING THE ANALYTICAL AND SYNTHETICAL METHODS OF DEMONSTRATION, DESIGNED FOR SCHOOLS AND ACADEMIES. BY JOHN M'NEVIN, TEACHBR OP BNOLISH AND MATHEMATICS BALTIMORE: PUBLISHED BY FIELDING LUCAS, JR. No. 138 Market street. PRINTED BY JAMES YOUNG, 1841. Price One Dollar. Entered according to Act of Congress, in the year 1840, by JOHN M'NEVIN, in the Clcrk'i Office, of the I land. BDUCATI01I libs; CONTENTS. (JA 102. I r.i... LECTCRE I. ration 12 Preliminary Remarks 13 lecture n. A.lditiou 18 LECTCRE HI. Multiplication 1 1 J Operations 37 Calculation of Bills 46 LECTCRE IV. Subtraction 54 LECTCRE V. Division 61 f Numbers 79 Analysis of Numbers 84 LECTURE VI. Decimals 90 Addition of Decimals 93 Multiplication of Decimals 94 Bebtraction of Decimals 96 i.ils 97 Reduction of Decimals 99 LECTURE VII. Induction of Weights, Mea- sures, &c 104 Addition of Compound Numbers. 129 Multiplication, .do do.. . .132 Subtraction of. .do.. . . . »do 1 •') I Dn ision of do. . . ^. .do. . . .136 LECTURE VIII. Fractions Simplified Preparatory Questions 140 n»al Fractions 143 NliOMl parts of a §.143 al Theorems in Fractions .111 Addition of Fractions 147 Reduction of mixed or Com- {KMUd Fraction* lib Page. Multiplication of Fractions 149 Subtraction of Fractions 151 Division of Fractions 153 Ratio of mixed Numbers Rule of Three in Fractions. . . .155 Invent Proportion by Fractions 156 LECTURE IX. i Proportion 1 37 Contracted Operations Inverse Proportion 169 Compound Proportion 163 Practical Arithmetic 164 How to Mark Goods * 168 Custom House Calculations 170 Tare and Tret 173 Interest 176 Method of Calculation in Bank- ing Institutions 178 Partial Payments on Bills at interest 183 Miscellaneous Questions 186 Insurance 188 Marine Insurance 188 i ance on Real Estate 188 Merchandise, &c 189 Commission 190 Discount 191 Buying and Selling Stocks 192 Barter. 193 Loss And Gain 195 Exercise Questions 199 • any Accounts 200 Apportioning the effects of a Bankrupt » MM Equation of Payments 203 General Averages 205 How to Average personal ac- counts ..207 net M377Q: h CONTEXTS. ftfft Exchange 209 Value of Gold Coins 210 European Coins reduced to dol- lars and cents 211 Mercantile Forms 212 Sterling money reduced to dol- lars and centi Exchange above or below par. .223 Negotiating Bills of Exchange. 225 ;ar Exchange 226 Involution 227 Extraction of the Square Root. .228 To find the distance that an ob- ject may be seen at sea 230 Extraction of the Cube Root bsffJWl principles assumed. 233 Single Position. . .235 Double Position 236 Annuities.. ...238 Annuities at Compound Intercst.239 Annuities in Reversion Perpetuitie* Page. Discount by Compound Interest. 245 Alligation 246 Arithmetical Progression 247 Geometrical Progression Duodecimals Mensuration important to Ship Builders To find Ship's Tonnage 254 Board or Lumber Measure 254 To Measure Scantling or Joist. .255 Papering Rooms 256 Carpenter's work 256 Bricklayer's work M a —ra tion of Superlkcs and Solids Guaging ... Abstract of mechanics Mechanical powers 267 On the Steam Engine . General Theorems Questions for Exercise RECOMMENDATION The following is from the pen of Richard Cotter, Esq., formerly Professor of Mathematics and Natural Philoso* />/iy, in Baltimore College. I have carefully examined Mr. M'Nevins' manuscript of the "United States Calculator," and am happy to assert, that the various rules contained therein have been beautifully arranged, and that it contains a variety of matter useful to the student and accountant, not to be found in any Arithmetical work extant. Under these cir- cumstances, I have no hesitation in saying, that it is a ■rork of superior merit. E. COTTER. DEDICATION. To the worthy instructers in the United States: — (.i:\iiimen — Permit me to present to your con- ation the following system of Arithmetic which is intended as another auxiliary to your pupils in the useful department of mercantile education. The plan of teach- ing exhibited in the following sheets being theory and practice combined, is different from all which have hither- to been published, and can be communicated by the principal to his pupil with that success which experience alone, will sufficiently demonstrate. The work is in dol- lars and cents, adapted to all sorts of business, and nothing appears obscure or mysterious, but all is made plain and I remain gentlemen, yours respectfully, JOHN M'NEVIN. Balto. Jan. 1, 1841. PREFACE When I entered upon the arduous task of simplifying a science of so much importance to the civilized world as Arithmetic, in every department of society, I was aware, that it would be attended with no inconsiderable labour, deep thinking and research: How far I have suc- ceeded to prove this work an auxiliary, to increase the general stock of knowledge already disseminated, (for the youth of our country) a generous, enlightened, and im- partial public will determine. At all events, it is the result of much experience, during twenty-five years teach- ing, and it is the opinion of learned gentlemen to whom the manuscript has been submitted for inspection, that if it be introduced into our schools and academies, it will be found to answer admirably, the end, for which it is in- tended, and, thereby prove to be a guide and a text book to the merchant, mechanic and man of business, and, a standard system of Arithmetic in the United States. Some persons may think, that books of a similar nature have been sufficiently multiplied; but in answer to this, I may with propriety state, that the only correct evidence of a spirit of improvement in our seminaries of learning, and that education is in a high state of prosperity, is, the con- tinually increasing demand for new school books which exists in every section of our country. I ask then, what friend of our numerous, and flourishing literary institutions can, for a moment wish to check the progress of improve- ment in our books and systems of instruction; — Indeed Mil PREFACE. it may with jh fety be mentioned, that when instructers become wholly content with the elements and principles which have long been in use the progress of improvement ! y at an end. a matter of fact that dispatch in business in every department, is no small accompli>hm< ■: i fore, to prepare pupils in the shortest period of time possible for business, and to afford assistance to teachers in the arduous and important work of education were the motives which led me to this pioduc m. I. I me ask, why is it that boys of common capacity can leam any kind of mechan- ical business in three or four years, when, in fact, double the time has been devoted to the attainment of only a nowledge of the rudiments of Arithm< How is this, I pause for a reply? In the systems generally used in the schools of our country, the questions in the different rules are li mixed" up with examples of Pounds, Shillings and Pence. Certainly this is wrong!! Because the very idea of pounds, shillings and pence is foreign to American students. Not only this, but the derangement of the rules, and the old -J protracted methods of calculation which are adopted in these old Arithmc t retard the progress of pupils and cause a dislike to their studies. This is a fact, which cannot be controverted. And, that two-thirds of the time usually spent in ac- quiring a knowledge of this useful art may be saved, by means of a proper system of Arithmetic and a correct mode of instruction is also, a fact well known to every judicious instructer. In no department of mathematical science have I been more deeply interested than in this, and hence, the reason, (as before mentioned) that I have laboured assiduously in preparing and collating a work, to suit the wants of the American people. In relation to foreign currencies, sterling money ,axL&\he currencies of other countries of Europe are treated of in Exchange: this is precisely as it should be: because, we PREFACE. IX that young persons are not lovers of long or hard ns and that they can learn with pleasure, when neither their memories are overcharged nor their under- standings put on the rack, from the use of tedious and ambiguous examples, which are interspersed throughout ins of Arithmetic said to be designed for American schools. In consequence of this, let us try to remedy this evil: In the first place, is it not well known, that the United States have political and civil institutions of their own, ami can these be upheld unless, our children are t to understand them by books and other means of instruction perfectly suited to the genius and constitution be country? Again, a proper system of Arithmetic in dollars and cents, is much wanted, in order that pupils may receive from their teachers ample satisfaction for their time and money. For it is manifest, that all, that concerns our public happiness, our union and peace with- in ourselves, all which tends to develope our resources, improve and perpetuate our institutions, all which may gives us wealth, strength, and glory among nations, depends on a general and systematic course of instruction in our schools. How ready,short, easy and familiar this work may be for actual performance, will appear by inspection. In conclusion, I cheerfully tender my most grateful acknowledgments to Richard Cotter, Esq. Professor of Mathematics and Natural Philosophy of thi^ ably and faithfully acquitted himself in its iination, and also, for the useful hints he has kindly suggested to me, which I conceive to be of utile im- portance to the work. With these remarks, and an ardent wish, that the States Calculator; or, Arithmetic Sim- plified" may be useful to the youth of my country, I cheerfully submit it to the patronage of an enlightened and it. :.t public. JOHN M'NEVIN. B alto. Jan. 1, 1811. UNITED STATES CALCULATOR; ARITHMETIC SIMPLIFIED ARITHMETIC is that part of the Mathematics which teaches the art of computation by numbers. All opera- tions in Arithmetic are performed by means of the fol- lowing figures: Cipher One Two Three Four Five Six Seven Eight Nine 12345678 9. There are five principal rules in Arithmetic, viz: No- tation and Numeration may be included in 1, Addition 2, Multiplication 3, Subtraction 4, Division 5. The proper definitions are inserted at the head of each rule. The Arabic numbers, are represented by the characters above written. The Roman letters also express number without limita- tion, but are chiefly used to mark dates, and the chapters, and sections of books. The following letters are used to express number. I V X L C D M 1 5 10 50 100 500 1000 1 ! ARITHMETIC. NUMERATION TABLE. 1 i ^ o.2 w "3 a sp a o " • * s • — ■/. -2-° 3 *3 S fl o.2 § 9*8:3 § eq »i **o 2 5 1 to w • — j3 * s« ? s^ 1^3^ |°=|" 2 Istla? sa?1a§1a?las I- 456 789 987 654 321 234321 Any number of figures whatever, may be accurately numerated by dividing them into periods of six figures each as in the above table, repeating Millions of Millions for billions, millions of millions of millions for trillions, .is often as the figures 2, 3, 4, &c. placed underneath in the table designates. Archimedes said "give me a place to stand and I shall remove the earth." The probable weight which Archimedes would have to raise is expressed by the fol- lowing number of pounds, viz: Trillions, Billions, Millions, lbs. 400,000,000,000,000,000,000,000 How many? Ans. 400,000 trillions of lbs. X NUMERATION. 13 LECTURE I. PRELIMINARY REMARKS. To use the language of "Lacroix," the idea of Num~ the latest and most difficult to form; because, before mind can arrive at such an abstract conception, it must be familiar with that process of classification, by which we successively remount from individuals to species, from species to genera and from genera to orders. The ge is lost in his attempts at numeration, and signifi- y expresses his inability to proceed by holding up \dkd fingers or pointing to the hairs of his Nature has furnished the great and universal standard lor computation in the fingers of the hand. All nations have accordingly reckoned by fives, and some barbarous tribes have scarcely advanced any further. After the fingers of one hand had been counted once, it was a second, and perhaps, a distant step, to proceed to those of the other. The primitive words expressing numbers, did not probably exceed jive. They counted by "tally" on wood, as the returns of our elections are made out on paper, thus |H| for five, making four marks in a straight line, and drawing a diagonal line for the fifth. To denote, six, seven, eight and nine, the North American Indians repeat the five, with the successive addition of one, two, three, and four. Could we trace the descent and affinity of the abbreviated terms denoting the num- bers from five to ten, it seems highly probable, that we should discover a similar process to have taken place in the formation of the most refined languages. The tea 2 14- NUMERATION - . digits of both hands being reckoned up, it then became necessary to repeat the operation. Such is the foundation of our Decimal Scale of Arithmetic Language still betrays, by its structure, the original mode of proceeding. To express the numbers beyond ten, the Laplanders com- bine an ordinal with a cardinal digit, thus eleven, twelve, &c. they denominate one ten and one, one ten and two, &c.; and in like manner, they call twenty-one, twenty-two, &c. two tens and one, two tens and two, &c. which they express by holding up both hands twice for twenty, three times for thirty, four times for forty, and so on. When they want to express forty-three, or forty-four, it will be four tens and three, four tens and four, &c. Our term eleven is supposed to be derived from the Saxon tin or one and liben to remain, and to signify, one leave or set aside ten. Twelve is of the like derivation, and means two laying aside ten. The same idea is suggested by our termination ty in the words twenty, thirty, #c. This syllable altogether distinct from ten, is derived from Ziehen to draw, and the meaning of twenty is strictly speaking two drawings, that is, the hands have been twice closed and the fingers counted over. , After ten was firmly es- tablished as the standard of numeration, it seemed, the most easy and consistent to proceed, by the same repeated composition. Both hands being closed, ten times would carry the reckoning up to a hundred. This word, original- ly hund, is of uncertain derivation; but the word thousand which occurs at the next stage of the progress, or the hundred added ten times, is clearly traced out, being only a contraction of duis hund or twice hundred, that is, the repetition or collection of hundreds. NUMERATION. Definition. — The art of numbering, is that, which teaches the notation, and just method of reading numbers. Number, presents itself, in two different forms; first, NUMERATION. 15 as a whole, without distinction of parts, as, when we speak of distance; as, the length of a line, from one extre- mity to the other. Secondly, number is designated as a collection of things or parts, and one of these things or parts is called a unit, hence, the units are expressed, by one, two, three, four, five, six, seven, eight, nine. The collections of ten Pt or tens by 10, 20, 30, 40, 50, 60, 70, 80. The collections of 10 tens or hundreds are expressed from names borrowed from the units, thus we say one hundred, two hundred, nine hundred, nine thousand, ten thousand, twenty thousand, one hundred thousand, two hundred thousand, &c. The collections of ten hundred thousand or of thousands of thousands, take the name of mi //ions, and are distinguished like the collection of thou- sands. J . Write in figures — Three hundred and eighty-six. I . II! No. 1. 38 6 2. Write in figures — Nine thousand three hundred and seventy-three. Hit No. 2. 9 3 7 3 *. J J'// A in figures — Forty-three thousand four hun- and eighteen. No. 3. 16 M MERATIOX. 4. One hundred and thirty-four thousand six hundred and thirty. 4 I M | I fltil % I 2 J 2 S No. 4. 13 4 6 3 5. One million three hundred and eighty-six thousand four hundred and twenty-six. till s 1 a i "5 . "2 «s a I J S J J ! 9 No. 5. 13 8 6 4 2 6 6. Nineteen millions one hundred and seventy-four thousand eight hundred and seven. 4 I 4 I i . I 12 il I 1 1 1 1 I 1 No. 6. 19 1 7 4 8 7 7. Six hundred and fifty millions four thousand five hundred and five. h 1 I I i Jl S § No. 7. 650 4 5 5 NUMERATION. 17 & Six thousand five hundred and thirteen millions two hundred and eighty thousand three hundred and fourteen. s iff! 1 I ill i I No. 8. 6513 2 8 3 14. When a number is written in figures, in enunciating or expressing it in language, it is necessary, to substitute for each of the figures, the word which it represents. The following example will illustrate this: — '5 5 5 3 13 * s 1 * I o «s 3 'si llll! a " = i5o|Co. 2 |og- . I ! 1 1 1 1 1 1 1 1 1 1 24897321580346 It is evident that the formation of numbers by the suc- cessive union of units, is independent of the units, as ap- pears from the above table, by means of which, we are e nabled to compound and decompound numbers, which is called calculation. We shall now explain the princi- pal rules, for the calculation of numbers, without regard to the nature of their units, and proceed to Addition. Q» I s ADDITION. LECTURE II. Let us now, review this lecture, and enquire, what does the conjunction and mean, when we say 4 and 5 make 9? A. Addition. Q. What then is addition? A. The collecting or putting together of numbi u Illustration. — 6 and 8 are 14, which is called the sum. Q. What is the sum called? A. The amount or total aggregate. Q. W r hat difference is there between the sum and the AMOl A. The sum means one particular sum, as 6 dollars, 8 dollars, or 10 dollars. And the amount is the sum total. These have another meaning when we speak of interest. See Interest, lecture 10th. ON ADDITION. (Sign +) It is a well known principle in Mental Philosophy, that the mind derives all its primary ideas from the immediate perception of the senses. A knowledge of this fact shows the importance and utility of employing sensible or tangible objects, in order to assist the juvenile capacity in comprehending the na- ture and combinations of abstract numbers. Nor should this be considered as a modern innovation, for we find, that the ancients frequently, had recourse to similar methods, when they attempted to teach the prin- ciples of Geometry. The Syrians, for instance, and most of the contiguous nations, looked to the Egyptians as teachers in most of the sciences. And it evidently ap- pears, that their instructions and communications were imparted, chiefly, through the medium of hieroglyphics. ADDITION*. 19 . and other figurative representations, adapted to and circumstances. There is one important quality of the mind which de- serves to be particularly noticed; for instance, when we speak of dollar. u< mean numerical increase, that is to say, addii cause, to add, is to increase from an unit. thousand of these American coins may be cast in aine die, and to all appearance may be precisely alike, neither does the mind conceive them under any idea of variety, but mere increase of number; from this principle, it may be laid down philosophically, that no individual can make great advances in intellectual improvement, beyond the immediate perception of sensible objects, such as pertain, to the five corporeal senses, seeing, hearing, g t tasting, and smelling. This is a doctrine, to which our general train of reason- ing will refer: for it is certain, there is no possibility of transmiting our ideas, but by reference, to sensible objects. The science of Arithmetic in its present structure, can be reduced to mathematical precision, its elements can be analyzed, its combinations discovered, and a basis laid for gradual improvement, according to its general and consis- tent laws, by Demonstrations, and Illustrations. Therefore, as the cultivation of the mind depends, chiefly on a correct habit of thinking, the teacher should, by all means convey his ideas in a plain and familiar manner, ;»nd in the commencement, have recourse to visible objects. 20 ADDITION. For instance, the diagram here laid down is a plan of an orchard, there are ten rows of apple trees in it, and five trees in each row, the question is, how many trees in the orchard? Ans. 50. OPERATION. 1st line I 2nd row 3rd do. 4th 5th ?th Mh 9th 10th do. do. do. do. do. do. do. 1" 10 10 10 10 M$L .. y >- ± x 1 -i. 50 Ans. Again, we next introduce, the system of addition by means of the Circular Digits. The teacher may com- mence with A in the class to count, in the following manner. 2 4 6 8 10 12 14 16 18 20 22 24 222 2 22222222 4 6 8 10 12 14 16 18 20 22 24 26 26 28 30 32 34 36 38 40 42 44 46 48 2222222222 22 28 30 32 34 36 38 40 42 44 46 48 50 or differently, according to the numbers on the Circular Digits above mentioned. Then, let the class commence with 3, 3+3=6. 6 9 12 15 18 21 24 27, &c. up to 51. 33333333 9 12 15 18 21 24 27 30 Then again, with 4, 4 and 4 are 8, 8 12 16 20 24 28 32 36, &c. up to 52. 44444444 12 16 20 24 28 32 36 40 ADDITION. 21 in small additions, pupils can readily i r in mental operations. When the addition class is called up, each scholar should be furnished with a card, numbered agreeably to the lessons in the Addition Table. 1st card numbered the same as les- son first; 2d card numbered the same as lesson second, and so on. Hence, it is evident, that when, the subject of ad- dition, is commenced, the pupils become highly delighted with this instructive and amusing process; because their ideas become interchanged, between one percipient thing and another, which, no doubt, emanate, from the use of these tangible objec The astonishing number, of changes, capable of being produced, by the various combinations of these elementary materials, afford, so many instances, of the ingenuity of man. The nine figures, with a cipher 0. give all possible pow- ers of numbers. A*« The figure A, B, C, D, is called a magic square. C" 8 3 4 1 S 9 6 7 2 KB The learner will find by adding the figures laterally, diagonally, and vertically, that the sum is 15. It appears by inspection, that the central figure 5 is a Magic number, and the favourite figure in nature, for all the machinery on earth is reducible to the following 5 distinctive forms, viz: the Wedge, Lever, Pulley, Wheel, and the Screw. ADDITION. vefive fingers and five toes on each hand and loot, and we have, also, five senses, as before mention- ed: — We find, that the ancients counted by tally on wood, the same as the returns of our elections are made out on paper 1>\ iivks, and, the North American Indians count, by holding up their hands, so, that, nature has fur- nished, the great and universal standard, of addition by 5. Addition is the first step to Mathematical knowledge, and is founded, on the first axiom, in Euclid, viz: the the whole is equal to its parts. MENIAL EXERCISES. How many are 5 and 1 ' How many are 8 and 5? " « " 6 and 8? " " 9 and 4? " " " 6 and 3? " " " 7 and 5? " " " 7 and 1 I " " " 8 and 9? " " " 9 and 71 " " " 9 and 9? The learned instructer may use his own discretion, on the subject of proposing questions. Q. Why do we carry, 1 for every 10 in addition? A . The reason is manifest, from numeration, we find, that numbers increase in a ten fold ratio. Example. — Gave $26 for a cow, and $9 for a load of hay, what did I give for both? Ans. $35 00. Operation, 2 6 Here 9 and 6 added together make 9 15, which is 5 above 10. Ans. 35 m ^ I set down 5, in the units place, and carry 1 to 2 which make 3, which I set down in the ten's place and find the sum is 35. Definition. — To add, is to collect together, to make firtt Lesson. 1 1 are 3 2 and 2 are 4 2 and 3 are 5 Second Lesson. 2 and 4 are 6 2 " 5 " 7 2 " 6 " 8 2 " 7 " 9 T/iird Luton. 2 and 8 are 10 2 " 9 « 1 1 2 " 10 " 12 2 " 11 " 13 2 " 12 « 14. Fourth Lesson. 5 and 1 are 6 5 " 2 " 7 5 « 3 " 8 5 " 4 " 9 6 " 5 " 10 5 " 6 " 11 Fifth Lesson. 5 and 7 are 12 5 " 8 " 13 5 " 9 " 14 5 " 10 " 15 5 « 11 " 16 5 « 12 " 17 Sixth Lesson. 3 and 1 are 4 3 " 2 « 5 3 " 3 " 6 a " 4 « 7 3 " 5 « 8 3 " 6 " 9 3 " 7 " 10 SevcniA Lesson. 3 and 8 are 11 3 " 9 " 12 3 " 10 " 13 3 "11 <• M 3 " 12 M 15 ADDITION. 2 )DITION TABLE Eighth Lesson. Fourteenth lesson. 6 and 1 are 7 8 and 1 are 9 6 " 2 " 8 8 " 2 " 10* 6 " 3 " 9 8 « 3 " 11 6 " 4 " 10 8 " 4 " 12 6 « 5 "11 8 " 5 " 13 6 " 6 k - rJ 8 " 6 « 14 A*tn/A Lesson. Fifteenth Lesson. 6 and 7 an 8 and 7 are 1 5 6 " 8 " 14 8 " 8 « 16 6 " 9 « 15 8 " 9 " 17 6 "10 " 16 8 " 10 « 18 6 " 11 " 17 8 "11 " 10 6 « 12 M 18 8 " 12 " 20 Tcn/A Lesson. Sixteenth Lesson. 4 and 1 are 5 9 and 1 are 10 4 " 2 " 6 9 « 2 « 11 4 " 3 " 7 9 « 3 « 12 4 " 4 " 8 9 " 4 " 13 4 " 5 " 9 9 « 5 « 14 4 " 6 " 10 9 « 6 " 15 4 " 7 " 11 Seventeenth Lesson. Eleventh Lesson. 9 and 7 are 16 4 and 8 are 12 9 " 8 « 17 4 " 9 " 13 9 « 9 " 18 4 " 10 " 14 9 « 10 " 19 4 «11 « 15 9 "11 "20 4 " 12 " 16 9 "12 "21 Twelfth Lesson. Eighteenth Lesson. 7 and 1 are 8 10 andl are 11 7 « 2 " 9 10 " 2 " 12 7 " 3 " 10 10 " 3 " 13 7 " 4 " 11 10 " 4 " 14 7 " 5 « 12 10 " 5 " 15 7 " 6 « 13 10 " 6 " 16 Thirteenth lAsson. Jfineteenth Lesson. 7 and 7 are 14 10 and 7 are 17 7 « 8 « 15 10 " 8 " 18 7 « 9 « 16 10 " 9 " 19 7 « 10 " 17 10 "10 "20 7 « 11 « 18 10 "11 "21 7 m 12 « 19 10 " 12 " 22 24. ADDITION. Twentieth Lesson. 1 1 and 1 are 12 Twenty-second Lesson. IS and 1 are 13 Twenty-fourth Lesson. 13 and 1 are 14 11 " 2 " 13 12 " 2 " 14 13 " 2 " 15 11 " 3 "14 12 " 3 " 15 13 " 3 " 16 11 " 4 "15 12 " 4 " 16 13 " 4 " 17 11 " 5 " 16 12 " 5 "17 13 " 5 " 18 11 " 6 "17 12 " 6 " 18 13 " 6 " 19 Twenty-first Lesson. 1 1 and 7 are 1 8 Twenty-third Usson. 1 2 and 7 are 19 Twenty-fifth Lesson. 13 and 7 are 20 11 " 8 "19 12 " 8 " 20 13 " 8 "21 11 " 9 "20 12 " 9 "21 13 " 9 " 22 11 "10 " 21 12 " 10 " 22 13 " 10 " 23 11 "11 "22 12 "11 " 23 13 "11 " 24 1] "15 " 23 12 « 12 " 24 13 " 12 " 25 ADDITION. 25 DIVERTING EXERCISES IN ADDITION, calculated to instruct and amuse the learner. 1 | 2 | 4 | 8 | 16 | 32 3 | 3 | 5| 9|17|33 5 | 6 | 6 ] 10 | 18 | 34 7 | 7 | 7 j 11 | 19 | 35 9 | 10 | 12 | 12 | 20 j 36 11 | 11 | 13 | 13 | 21 | 37 13 | 14 | 14 | 14 | 22 | 33 15 | 15 | 15 | 15 | 23 | 39 17 | 18 | 20 | 24 | 21 | 40 19 | 19 | 21 | 25 | 25 | 41 21 | 22 | 22 | 26 | 26 | 42 23' | 23 | 23 | 27 J 27 | 43 25 | 26 | 28 | 18 | 28 | 44 27 | 27 | 29 | 29 | 29 | 45 29 | SO | 30 | SO | 30 | 46 31 | 31 ] SI | 31 | 31 | 47 33 j 34 J 36 | 40 | 48 | 48 35 | 35 | 37 | 41 | 49 j 49 37 | 38 | 38 | 42 | 50 { 50 39 | 39 | 39 | 43 | 51 | 51 41 | 42 | 44 | 44 | 53 | 52 43 | 43 | 45 J 45 | 53 | 53 45 | 46 | 46 | 46 | 54 j 54 47 | 47 | . 47 | 47 | 55 | 55 49 | 50 | 52 | 56 | 56 | 56 51 | 51 | 63 | 57 | 57 | 57 55 | 55 | 55 | 59 j 59 | 59 67 j 58 | 60 | 60 | 60 | 60 59 | 59 | 61 | 61 | 61 | 61 CI | 62 | 62 | 62 | 62 | 62 63 | 63 | 63 | 63 | 63 | 63 26 ADDITION. Elucidation of the foregoing Tail p. On the foregoing page you see, that 1, 2, 4, 8, 16 and 32 are in the 1st horizontal row and no where else; but 1 +2=3, shews that 3 is to be found in the 1st and 2nd columns, because it requires 1 and 2 to make 3. If it re- quires three or more of the top numbers to make one of the numbers in the table, as 1-f 2-f4-f 8=15, the num- ber 15 is to be found in all the columns in which 1, 2,&c. are found at top; and so it is with all the other numbers, which are generated in the same way. The horizontal rank of numbers, 1, 2, 4, 8, &c. are in geometrical ratio, and the perpendicular rank on the left in arithmetical ra- tio; both together the abscisae and ordinates of the Loga- rithmic curve. The rank of geometricals at the top viz: 1 . 2, 4, &c., afford such combinations as make up all oth- er numbers, in whatever column any number stands, the sum of the geometricals which stand over such a number is equal to it, and the reason, why one can guess any num- ber thought on, is, the sum of the numbers which stand over it, in the first horizontal row, makes it, and no other exactly. Suppose for example, I fix upon the number 50, and ask you to tell, which number I thought on, you would ask me in which of the columns, and how many does it stand in? I would say in the second and two last columns, you would then put 2+16+32=50, and tell me to a certainty; because 50 is no where else, but in the second and two last columns. Whatever number you guess upon, if only, in the first, second, third, fourth, fifth, or sixth columns, the numbers must be 1, 2, 4, 8, 16 or 32, because each of them are found but once. But where * it takes two or more of the figures in the horizontal row of geometricals, to make the one you think on, the number is to be found in all the columns, exactly under the geo- metricals, which compose it, and no where else, hence it is easy to understand the diverting exercises, if another column was added its head number would be 64, from ADDITION. 27 which a table could be generated from 1 to 127, because 63+64=127. M I ;XTAL EXERCISES. # 1. How many are 5 and 4? 6 and 8? 6 and 3? J. How many are 9 and 61 8 and 9? 12 and 8? Remark, in relation to the utility of proposing questions to be answered mentally in addition, the plan is very good, but in order to be expert in answering, the pupil should be required to recite the Addition Table. Mental exer- cises answer very well for small children, and for such, I would recommend " Colburn's First Lessons." 1 . Bought an orange for 3 cents, and nuts for 6 cents. How much did all cost? J. Richard has 10 cents and Robert has 4 cents. How many have both? Si You gave 12 cents for a knife, 4 cents for an ink- stand and 1 cent for a slate pencil. How much will all of them come I 4. If you have 4 nuts in one hand and 4 in the other. How many in both? 5. Thomas has 35 marbles in one pocket and 21 in the other. How many in both? The teacher can propose similar questions to suit the capacity of the pupils. DOLLARS AND CENTS. The addition of dollars and cents is the same as the ad- dition of whole numbers, with the exception of pointing out 2 figures, beginning at the units place for cents. doll*, cts. dolls, els. dolls, cts. Add 371.25 Add 9863.37 Add 16.96 648.75 4962.74 44.92 934.63 3248.64 84.73 $1954.63 28 ADDITION. doll*. cts. doll*, eta. dolls, cts 1 1349.16 Add 64669.74 Add 4.99 % 4469.74 1*242.68 6.66 9324.68 93646.95 9.87 6466.67 44668.67 4.63 9464.63 APPLICATION. 1. A gentleman gave $17.50 cents for a coat, $12 for pantaloons, $5.50 for a hat; what did they all come to? Ans. $35. V merchant sent out his clerk to collect money; he collected $50 from one person, from a second $14.94, from a third $53.35, and fourth $31.43 cts.; how many dollars did he collect in all? Ans. $149.72 cts. 1 . How many half cents in a cent? Ans. 2. 2. How many quarter cents in a cent? Ans. 4. 3. How many quarter cents in half a cent? Ans. 2. ILLUSTRATION. half or 2 quarters. 1 quarter. ADDITION. doll*, cts. dolls, cts. dolls, cts. Add 376.18* Add 44.37* Add 16.42* •Jik36* 22.12* 17.96* 320.55 dolls, els. dolls, cts. Add 67942.37* Add 6474.62* 42464,'J 4248.68* 49644;! 4149.371 dolls, cti. dolls, cts. Add 6426846.37* Add 33642.62* 4247494.185 94696.47* J1692.62* 92469.621 dolls, cts. dolls, cts. Add 64946.931 Add 924867.62* 44924.3 li 846493.681 94246.62* 924696.37* 84647.37* 448693.3H 29 PROMISCUOUS EXAMPLES. 1 . I owe a merchant $7.00 for sugar, $5 for coffee, tor flour, $3 for tea, and $2 for salt. What is the whole sum? Ans. 34 dollars. A merchant bought 4 pieces domestic muslin, as follows; No. 1 contained 28 yards, for $2.80; No. 2 con- d 29 yards for $3.19 cents; No. 3 contained 28* yards, cost $3.13*; No. 4, 27* yards, co«t $3.02* cents. How many yards in the whole, and what did they all come to? J<k. dolls, cts. >. 1. 28 2.80 2. 29 3.19 3. 28* 3.13* 4. 27* 3.02* V 30 ADDITION. A Butcher bought of one man 25 head of cattle for $450.87$, of another 15 head for $200.75, and of an- other 9 head for $77.62 J. How many did he buy and what did the whole come to? Ans. 59 head, whole cost $729.25. ADDITION, OR AGREEABLYTO A MERCANTILE PHRASE, "FOOTING" OF DOLLARS AND CENTS. A owes to one creditor 569 dollars, to another 3961, to another 581, to another 6116, to another 469, to ano- ther 506, to another 69381, and, to another 1261; what does he owe them all? Ans. $82871.00 The mode of footing here introduced, has been adopted in the Bank of the United States some years ago, and, of course, has been used by the best Book-keepers in the Union, on account of the plan, being not only simple, but ingenious. OPERATION. Place units under units, tens under tens, & hun- dreds under hundreds &C.&C. 596 3961 581 6116 469 506 69381 1261 9 1 1 1st column. ft 7 2nd m 9 1 8 3rd a 2 2 4th U - 1 8 5th u 82871 as before. 82871 ILLUSTRATION. Place the addition of every column of figures to be add- ed, under each other. Then when set down, reject the ten's place in every line except the last line. EXAMPLE. 3 The ten's place being cut off, or rejected, in every line ADDITION. 31 except the last, there remains 82871. The lower figure 8 being set down first, 2 next, 8 next, 7 next, and 1 next; which written, will appear as above, 82871. dolls, cts. Add 4168.34 7816.56 1.46 k34 6813.49 $37036.19 As before, add up every column in succession, which place under each other, and reject or cut off the ten's place, and you have a correct addition in one line. EXAMPLE 2. itioi i of the 1st column is 2 9 a u " 2nd u 2 1 U U " 3rd a 2 6 u a " 4th a 1 3 u a " 5th a 3 u U " 6th u 37 $37036.19 ANOTHER METHOD. By adding up, every line, simply without carrying to the units place, but "skipping" one figure back, every time, until you get through. A sufficient proof in itself, that units must be placed under units, tens under tens, hundreds under hundreds, and thousands under thou- sands, &c, or cents under cents, and dollars under dol- lars, in the same order, thus: Addition of first line, 29 a " '2nd 19 u " 3rd, 24 (« " 4th, 11 M « 5th, 29 u " 6th, 34 $37036.19 32 ADDITION. 3. A man owes Ezekiel Dorsey $6374.18 cts.; to B. M'Donough $4397.65 cts.; to John E. Stansbury $9365 94 cents, and to the Merchants' Bank $7986.44 cents; how much does he owe all? Ans. $28124.21. 4. A man bought a barrel of flour for $5.75, a barrel of molasses for $29.00, and a barrel of rum for $36.00; bow much did he pay for all the articles. Ans. $70.75. 5. From the creation of the world to the flood, was 1656 years; from thence to the building of Solomon's Temple, 1340 years; thence to the birth of our Saviour, 1008 years. In what year of the world was the birth of Christ? Answer. Anno Mundi, (in the year of the world) 4004 years. 1 A PENSIVE ADDITIONS, <l"!ls. cts. EXAMPLE. Add 148697.69 Addition of the 1st line is 5 o 448644. 1 1 " 2nd " 5 4 934444.78 " 3rd " 4 5 .945464.64 " 4th " 6 'A 948492.69 « 5th « 4 4 947498.75 " 6th « 5 7 347698.63 " 7th " 3 7 456493.78 a _ " 8th " 57 In rejecting the ten's place in every line except the last, the $5777435.10 dolls, cts. Add 64668.16 remainder is 57774354, which 94648.74 is set down for the answer. 94844.65 dolls, cts. 47486.94 Add 64967.63 9464 94648.62 94648.63 94738,64 94646.68 Add 94666846.63 94484.62 46494646.74 94646468.64 Add 96486.63 54646462.63 46486.64 93287.64 64968.69 69696.87 69486.63 ADDITION. 33 Q. How are we to place figures in addition? A. Units, under units, tens under tens, hundreds un- der hundreds, thousands under thousands, &c. dolls, cts. dolls, cts. Add 6469.86 Add 6469.64 !»,87 9469.77 4698.75 9469.78 9468.69 9464.63 6469.68 9448.97 9464.99 9864.63 6469 68 M64.77 dolls. cts. Add 64698.69 46949.64 dolls, cts. Add 6*987*6.68 94674.69 6469696.75 24.64 9446946.68 94646.69 94c;:i "!'■"•> 94698.84 9646975.44 92465.75 31 MULTIPLICATION. LECTURE III. ON MULTIPLICATION. (Sign x.) Q. What i.s multiplication? A. Multiplication is a concise way of performing ex- tensive additions; because, to multiply, is to increase one number by another, as often, as there are unr that number, by which the one is increased. Q. How many parts has it? A. Three; the Multiplier, Multiplicand and the Rec- tangle or Product. Q. Which number is the multiplier? A. The number you multiply by. The reader will find, that the Multiplication Table, in this work, is different from the common form, and, that, the system, now, being introduced, for teaching the ele- ments of figures, in multiplication, leads to great improve- ment in mental researches. Inasmuch as, it unites The- ory with Practice; for, although, multiplication in itself, is simple, its minor variations are endless. Because, the relative value of figures, connected with the workings, of thought, adapt themselves, to every imaginable form, and run. into each other, by such nice gradations, as are only obvious, to the keenest observations of philosophy; and yet all this amazing system of intellectual improve- ment, depends, on the combinations of nine figures, with a cipher, called, the nine digits. Mt'LTlPLICATIOK. 35 MULTIPLICATION TABLE. 2 ce 1 IS 1! Li i: t I ; 9 2 2 4 • 8 10 12 14 |16 18 3 tniiod 1 2 3 3 3 4 3 5 3 6 3 7 3 8 3 9 3 3 6 1 » |12 [15 18 21 i 24 |*7 4 timet 1 1-3 3 4 4 4 5 4 6 4 7 4 8 4 9 4 4 1 8 |12 |lt l«> u | 28 | 32 | 36 j tinu-s 1 2 5 3 5 4 5 5 5 5 7 5 8 5 9 5 6 10 15 20 1 25 30 35 | 40 1 45 6 times 2 6 3 6 4 6 5 6 6 1 6 7 6 8 6 9 6 6 12 |18 |24 | 30 36 1 42 |48 | 54 7 limes 1 ! ! 4 7 5 7 6 7 7 7 8 7 9 7 1 14 21 1 28 | 35 1 42 1 49 | 56 | 63 8 times 1 2 8 3 8 4 8 5 8 6 8 7 8 8 8 9 8 8 16 24 32 40 48 56 64 Jl i timet l 2 9 3 9 4 9 5 9 6 1 8 9 9 9 • 18 27 36 45 54 63 |72 |81 1 Q. What is a multiple? A. It is a term in multiplication when one number, is contained in another several times; as, 15 is the mul- tiple of 3. Containing it 5 times. Q. What is the multiple of 24? A. 4, «,8, &c. Q. How can the product of two numbers be expressed in multiplication? A. As 3 X 5 = 15 product 36 MULTIPLICATION. Q. In this case, which is the multiplicand? A. 3. Q. Which is the multiplier? A. 5. Q. Can either be a multiplier? A. Yes? because 3x5«5x3=15as abo\ Mode of reciting the Table. — The learner should be taught to recite thus: beginning with 2ce — 2ce 1 are 2, 1 , or once, 2 are 2; 2 ce 2 are 4, and repeat, 2 ce 2 are 4; 2 ce 3 are 6, vice vers /. .'* times 2 are 6; 2 ce 4- are 8, 4 times 2 are 8; 2 ce 5 are 10, 5 times 2 are 10; 2 ce 6 are s 2 are 12; 2 ce 7 are 14, 7 times 2 are 14; 2 ce 8 are 16, 8 times 2 are 16; 2 ce 9 are 18, 9 times 2 . Again, begin with 3 times, 3 times 2 are 6, 2 ce 3 are 6; three times 3 are 9. Repeat, 3 times 3 are 9, 3 times 4 axe 12, 4 times three are 12, and so on, alternately. SYSTEM OF TEACHING. Let the teacher procure 8 sets of 'circular digits f as above represented, numbered agreeably to the form of the multiplication table. First lesson. Second lesson. 22345 6789 12222 2222 First set. 2 4 6 8 10 12 14 16 18 Second set. MULTIPLICATION. 37 Tkird Usson. Fourth Uston* 32345 6789 13333 3333 3 6 9 12 15 18 21 24 27 The circles to be numbered on both sides as above: Card No. 1, to shew lesson 1st on one side, and lesson 2d on the other. Card No. 2, to represent lesson 3d on one side, and lesson 4th on the other, and so on. By this plan, children five or six years of age, can be taught to multiply and answer mentally in a few days. CASE I. — Point two places of figures for cents, begin- n ing at the right hand, counting from right to left* 1. Sold 16 lbs. of Sugar at 6 cts. 14, « u « 7 a 12 " " " 6 " 10 " " " 8 " $3.46 Sold 8 lbs. of Sugar at 7 cts. 6 " " " 8 " 5 " « " 10 " 3 " " " 8 " I <t « « g M Whole amount, $5.33 3. Sold 3 yds. at $3.00 4 « « 5.00 9 " H 6.00 7 « " 9.00 $146.00 4. Sold 15 lbs. at 8 cts. 18 " " 10 " 16 « " 11 " 18 « " 12 « 19 « " 15 " $9.77 ■AS MULTIPLICATION'. 5. A lady bought the following items 13 yds. Calico at 1, &C. 22 cts. 8 " " u 32 " 5 M Domestic muslin (4 13 « 40 " Muslin || 17 « $12.87 6. Bo't 8 yds. Tape 10 " " at 5 cts. 8 « 11 " « u 9 « 12 " * a 8 " $3.15 7. Sold 10 lbs. of Coffee at 10 cts. 12 " " ii 9 « 15 " " N 12 " $3.88 8. Sold 16 yds. Muslin 12 « " at M 11 cts. 13 « 15 « " II 14 <{ 9 " " u 13 " 8 " " II 15 « $7.79 9. A lady bought the following articles, viz: 8 yds. Calico at 31 cts. 12 " Domestic muslin " 15 " 6 " Canton flannel H 20 " 3 " Ribband (C 15 £{ 9 " Bombazine M 44 " $9.89 10. A boy bought the following items, viz: 2 Slates at 27 cts. 3 Copy Books " 10 « 25 Quills " 2 « 100 Slate pencils for 20 " r $1.54 MULTIPLICATION. 39 11. Sold 16 yds. at 9 cts. 14 " " . 8 " 17 " M 9 " 16 " "11 " 17 u a 7 " 12. Sold 14 lbs. at 9 cts. 13 " " 8 " 12 " " 7 " $7.04 13 " " 6 14 « " 9 15 « " 8 13. Sold 13 lbs. at 5 cts. 16 " a 4 * 19 " tt 7 " 18 " tt 8 " 19 " a 6 " 24 « it 8 " $6.3$ $7.12 14. Sold 19 lbs. at 15 cts. 22 " " 16 •• 23 " "17 " 24 " u 18 " 32 « " 19 " 45 " " 21 " $30.13 15. Bought 10 pieces of muslin, each 29 yards, at 22 cents. What is the amount? Ans. $63.80. 16. Sold 5 bags of coffee each weighing 156 lbs. at ■uts. Ans. $117.00 17. Bought 125 barrels of flour at $4.81 per barrel. Ans. $601.25 18. Sold 125 cords wood at $6 pr. cord. Ans. $750.00 19. Sold 144 bbls. apples at 75c. per lb. Ans. $108.00 40 MULTIPLICATION. 20. Bought 162 bushels of oats at 44 cents. What is the amount? Ans. $71.28. 21. Sold 132 jards cassinett at 87 cents. What is $114.84 e amount? Sold 19 yds. at 37 cts, 22 " u 31 u 25 " M 42 u 28 " M 45 u 44 " (C 65 u $65.55 Sold to Charles B. Needles, 3 yds. green cloth at $5.25, per yd. $15.75 4 yds. Muslin at 35 cts. per yd. 1 .40 $17.15 CASE II. MENTAL OPERATIONS. 1. What will 3 oranges come to at 4 cents a piece? Ans. 12c. J. What will 6 quarts of cherries come to at 5 cents per quart? Ans. 30c. 3. What will 6 pounds of sugar come at 9 cents per pound? Ans. 54c. 4. What will 8 yards of ribbon come to 8 cents per yard? Ans. 64c. 5. If a man travel 3 miles an hour, how many miles will he travel in 5 hours? Ans. 15 <>. How many will he travel in 6 hours? Ans. 18 7. How many in 8 hours? Ans. 24 8. There are 12 inches in a foot, how many in 2 feet? Ans. 24 1 9. How many inches in 4 feet? Ans. 48 10. How many inches in 8 feet? Ans. 96 11. There are 3 feet in a yard. 12. How many feet in 7 yards? Ans. 21 13. How many feet in 9 yards? Ans. 27 MULTIPLICATION. 41 I 1. How many feet in 12 yards] Ana. 36 15. There are 7 da\ tek. How many days in Ans. 56 1 '-. How many days in 9 weeks? Ans. 63 17. A league is 3 miles, how many miles in 91 leagues? Ans. 273 18. What will 15 pounds of sugar come to at 10 cents? Ans. $1.50 The annexed diagram of an orchard is applicable in this case, to assist the learner in Multiplication* There are 10 rows of apple trees in it, and 5 trees in rach row. The question is, how many trees in the or- chard? Ans. 50 Here there are 10 rows, and 5 trees in each row. Multiply 10 By 5 Ans. 50 trees. From the above diagram we find that increase con- of .hid it ion and Multiplication. QUESTIONS FOR EXERCISE. 1. Calculate the amount of 126 lbs. butter at 15 cents per pound? $18.90 Calculate the amount of 19 cords of wood at $ »rd? *71.2f> 3. 169 bbls. cider at $2.12 per barrel? $358.28 4. 63 lbs. tea at $.93 per pound? $58.59 i ^9 lbs. sole leather, at 19 cts. per lb? $149.91 6. 47 thousand shingles at $3.75 per M? $176.25 4 # 42 MULTIPLICATION. 7. 169 boxes of oranges at $6.71 per box] $1133.99 8. 17 score penknives at 17 cts. each? $57.80 9. 96 yds. broad cloth at $5.75 per yard? $552.00 10. 421 bushels wheat at $1.35 per bushel? $568.35 1 1 . Sold 6 bags of coffee per lb. as follows: No. 1 160 lbs. at 15 cts. per lb. $24.00 •J 130 « " 15 H 19.50 3 140 « " 14 a 19.60 4 120 « " 13 a 15.60 5 148 " a is (( L7.76 6 139 " " 11 a 15.29 $111.75 'Sums' in which there are Ciphers at the right hand of the Multiplier or Multiplicand. v.. — Cut off the ciphers and add them to the result when multiplied. 2(0 No. 2 8 ^60 No. 3 987(000 148(000 64(00 4(00 2560000 7896 3948 !»s7S 146076000000 Sums in which Ciphers are intermixed. Rule. — Skip one place for every cipher. No. 1 106 No. 2 120013 204 408 No. 3 374957 2009 424 •J 1 2 21624 No. 4 978415 No. 5 526228 50208 400026 MULTIPLICATION. 43 No. 6 40085 3024 No. 7 4240 3109 I in wh ich Ciphers are intermixed with the figures of the Multiplier. 1. A merchant bought 106 hhds. of wine at $204 per hhd. How many dollars did he pay for all? Multiplicand, Multiplier, 106 204 424 212 21624 Product. No. 21 20013 No. 3 408 347957 No. 4 978415 2009 50208 No. 5 526228 70016 No. 6 356984 400026 Sums in which the Multiplier is 10, 100, 1000, 8Cc. 1 . A merchant bought 76 barrels of flour at $10 per barrel. How many dollars did he pay for the whole? Itiplicand, - - 76 Multiplier, - 10 Product, 43128 No. 3 10 760 8653 100 No. 4 213 1000 29187 No. 6 8674 No. 7 435000 10000 100000 9350000 EXERCISE QUESTIONS. 1. If you pay $7 for a barrel of flour, how many dol- lars must you pay for 125 bar Ans. $875.00 2. If you pay 8 cents for one lb. of pork, how many cent* must you pay for 375 pounds? Ans. $30.00 44- MULTIPLICATION. 3. A merchant bought 225 lemons at 5 cents a piece, how many cents did he pay for the whole? Ans. $1 1.25 4. James has 125 nuts, and William has 5 times as many, how many nuts has William? Ans. 625 5. A merchant has 9 boxes of raisins, each containing 15 pounds, how many pounds do all the boxes contain? Ans. 135 lbs. 6. A farmer sold 428 pounds of cheese at 7 cents a pound, how many cents did he receive for the whole? Ans. $29.96 7. If you pay $7 for one yard of broad cloth, how many dollars must you pay for 6 yards? Ans. $42.00 8. If the wages of one man for a year be $132, what u ill the wages of 12 men be? Ans. $1584.00 9. There are 24 hours in a day, and 7 days in a week, how many hours in one week? Ans. 168 10. There are in a hhd. 63 gallons, how many gallons I- hogsheads? Ans. 1512 1 1 . To the Senate of the United States each of the 26 States, sends two members, how many members are there in the Senate of the United States? Ans. 52 1 2. If a steamboat run 15 miles in one hour, how many miles can it run in 24 hours! Ans. 360 13. In 20 years how many days, allowing a year to be 365 days? Ans. 7300 14. What will 168 reams of medium paper come to at $4.25 per ream? Ans. $714.00 15. Bought 120 boxes of oranges at $5.35 per box. Ans. $642.00 16. 148 yds. broad cloth at $3.75 pr. yd. Ans. $555.00 17. 95 yds. cassinett at 62$ cts. pr. yd. Ans. $59.37 A 18. 144 yds. of broad cloth at $3.95 pr. yd. $568.80 19. James Young, Bought of John Wilson, 12 lbs. loaf sugar, at 17 cts. 10 " brown do. " 9 " 12 " coffee, " 15 " Received payment, $4.74 MULTIPLICATION. 45 QUESTIONS FOR THE SLATE. 1. What will 14 lbs. sugar come to at 13 cents per lb? ^ 2. 16 yds. at 17 " per yd. 3. 19 " << 15 " " 4. 64 " M 55 " " 5. 83 " M 86 « " $113.97 Sold to John R. Kemp, 6. 74 gals, of ' wine at 72 cts. per gal. 7. 98 « « 91 " " 8. 43 square feet of boards at 45 cts. per foot. 9. 54 bushs. of corn at 62 " per bush. 10. 85 « « u 87 « «' 11. 97 gallons of brandy at 95 " per gal. Amount, $361.39 Bought of David Simmons, 12. 23 yds. M 25 cts. per yard? 13. 28 " cc 29 " " 14. 36 " u 37 " u 1."). 48 " u 43 " « 16. 43 " u 48 « « $68.47 Bought of Ross Patterson, 17. 46 lbs. of coffee at 19 cts. per lb. 18. 49 yds. u 46 " per yd. 19. 54 " M 59 « « 20. 59 " u 54 4< u 21. 99 " u 92 " c< Amount, $186.08 109 yds. at $1.08 cts. per yard? 1. 117 " U 1.15 " « 119 bushels of wheat at 1.18 « per bush. 34 « of corn at 37 " " 26. 56 « << 54 " " 27. 125 « of wheat at 1.25 « " $591.76 k> MULTIPLICATION. 28. What will 28 yds. muslin come to at 22 cts. pr. yd? 29. 25 yds, 30. 17 « 31. 37 « 32, 33 " 33. 47 « 34. 48 « at « M II « Sold to George Harwood, 35. 77 lbs. of tea at 94 « « 97 gallons of wine u 36. 37. 38. 39. 40. 88 " « « 95 it u u 79 bushels of corn u 28 « 23 « 33 « 36 « 46 « 47 " $85.34 80 cts. Amount, per lb? 95 « " 93 * per gal. 87 « " 94 " " 76 * pr. bush. $552.35 CALCULATION OF BILLS. Sold John H. Sharp, 3 yds. broad cloth at 5 u brown holland <; 10 « padding « 18 « red flannel « Sold Oliver J. Griffin, 460 bales cotton net, 138,700 lbs, 210 « « 63,200 lbs. 250 « « 75,400 lbs. Sold Thomas R. Steele, 3 sugar loaves, wt. §1 lbs. at 19 lbs. of rice « 15 lbs. of pepper " 33 lbs. of cloves " $5 87 cts. per yard. 33 " « 62 " " 43 « * $33.20 at 24 cts. per lb. at 22 " « at 23 " " $64534.00 17 cts. per lb. 8 " " 23 « « 11 " " $17.27 MULTIPLICATION. 47 Sold William Hebden, 18 yds. calico at 37 cts. 34- " muslin " 15 " 63 « linen « 73 " 56 M flannel « 44 " 39 " cassimere " 87 " $116.32 Bought of Cornelius Ryan, 84 cords oak wood at $3.75 38 « hickory " 5.62 58 « yellow pine" 2.62 $680.52 Bought of Messrs. Wheelright, Turner, & Mudge, 64 reams of medium paper at $2.75 96 " " " " 3.25 84 " " " " 3.50 66 « " " " 4.25 $1062.50 Bought the following bill of goods, viz: 16 yds. domestic muslin at 15 cts. 12 «* " " " 18 " 19 « u m « 27 M 25 « a a tt 37 tt $17.04 Sold Samuel James, 12 yds. superfine cloth at $6.25 per yard. 3 pieces linen cont'ing 37 yds. at 60c. il $97.20 Bought of Henry Harwood, 3 yds. blue cloth at $4.25 cts. 14 « muslin « 11 " 8 skeins silk " 6 « $14.77 Sold James Young, 2 pieces Irish linen cont'ing 46 yds. at 75 cts. pr. yd. 3 " Canton flannel " 60 "13 " *• 1 « silk "36 "63 u " — $64.98 48 MULTIPLICATION. Sold Isaac Shirk, 9 bbls. superfine flour at $4.75 cts. per bbl g a u a u 5.25 " " 84 u u u u 5.33 « " 42 half bbls. « * a 5.28 " " $643.35 Sold G. W. Goddard, 1 bag Laguayra coffee, wt. 1 60 lbs. at 10 cts. pr lb. 1 « Java " wt. 144 lbs. at 11 « « 1 « Rio " wt 135 lbs. at 11 " " 1 " St. Domingo" wt. 120 lbs. at 10 « " — $58.69 Bought of J. S. Barry, 16 yds. cassinett at 75 cts. per yard. 84 « bombazine " 44 « 81 " flannel « 56 " « 36 « muslin « 14 " « — $99.36 13. Sold 49 green hides at $1.95 cents each, what did they come to? Ans. $95.55 BILLS. 14,. Philadelphia, January 3, 1840. Mr. Jacob Miller, Bought of P. Mercer, 12 yds. superfine cloth at $5.30 cts. per yard. 2 pieces linen cont'ing 37 yds. at 80 " " * $93.20 Baltimore, January 1, 1840. Mr. Joseph Ryan, Bought of P. Mercer, 6 silk cravats at $1.50 cts. each. 1 doz. pair kid gloves " 75 " 10 yds. black velvet « 2.60 « per yard. 3 red Merino shawls " 12.50 " each. $81.50 MULTIPLICATION. 49 MERCANTILE CALCULATIONS. Baltimore, June 4, 1840. Mr. Adam Cook, Bought of Job Lewis, 2 pieces Irish linen, cont'ing 46 yds. at $1.20 pr. yd. 3 « Canton flannel '• 95 yds. at 60 " 1 " silk « 30 yds. at 50 « $127.20 Mr. Isaac Williams, 17 yds. flannel 13 lbs. tea 47 u cheese Baltimore, January 3, 1840. Bought of Thomas Ferrel, at 45 cts. per yard. " 98 " per lb. ci 12 u '* — $26.03 Baltimore, June 5, 1840. Mr. Jonas Bailey, 19 yds. superfine cloth at 20 « linen " 30 " muslin " 2 black silk cravats il Bought of P. Mercer, $6.40 per yard. 1.12 " 15 « 1 40 each. $151.30 Baltimore, June 6, 1840. Mr. Job Lewis, Bought of P. Mercer, 37 yds. cassimere at $1.60, 40 " muslin « 17, A lot of trimmings. Amount per bill $27.43 A lot of ribbons. " *< 6.00 $99.43 19 yds. lace at $2.27 per yard. 14 " ribbon (< 18 " 24 " " u 25 « 13 fans u 13 each. $53.34 50 MULTIPLICATION. Mr. Thomas Clark, 12 yds. muslin 10 « 6 « calico drab cloth BILLS. Baltimore, June 6, 1840. Mr. John Abbott, Bought of S. Brewer, 17 pieces Irish linen, cont'ing216yds.at$1.10 pr. yd. 15 yds. book muslin at 56 M $246.00 Baltimore, June 7, 1840. Bought of Samuel Brown, at 38 cts. per yard. « 62 " " « $4.7 « $39.26 Baltimore, June 7, 1840. Bought of P. Fenby, at $1.25 per gal. " 15 per lb. " 14 " $35.19 Baltimore, July 7, 1840. Mr. Samuel A. W. Campbell, Bought of E. Dorsey, 3 M (thousand) quills at $7.50 4 doz. spelling books " 2.25 David Lusby, 25 gallons of rum 16 lbs. candles 11 " soap 1 ream of paper Mr. George Mitchell, 12 yds. muslin 10 " calico 3 " cassimere Mr. Wm. Kemp, 20 pieces of hanging paper at 8 " superior " 3.00 $34.50 Baltimore, June 7, 1840. Bought of John Andrews, at 20 cts. per yard. « 28 " " — $7.81 Baltimore, June 7, 1840. Bought of G. P. Knotts, 50 cts. 1.00 " $18.00 MULTIPLICATION. 51 Baltimore, June 8, 1840. Mr. Richard BrufT, Bought of E. Dorsey, 13 lbs. loaf sugar at 20 cts. per lb. 2 doz. cups and saucers " $2.75 per doz. 26 lbs. coffee " 15 cts. per lb. 36 " brown sugar " 10 " " 4 " Imperial tea « 1.25 " « 5 " mustard u 15 " " •' green tea " 87 « < $23.09 Baltimore, June 8, 1840. Mi. Nicholas King, Bought of E. Dorsey, 2 doz. knives and forks at $2.50 per doz. 3 pair sad irons " 1.25 i doz. waiters " 2.00 1 castor " 3.00 $13.75 Baltimore, June 8, 1840. Mr. James Chandler, Bought of Sam'l Brown, 1 piece of long lawn, 20 yds. at $1.12 per yd. 6 yds. linen cambric " 4.50 " $49.40 Baltimore, June 10, 1840. Mr. E. C. Johnson, Jr. Bought of Sam'l Brown, 2 pieces American nankeen, 11 yds. at 31 cts. pr. yd. 2 pair cotton hose at 50 " pr. pr. 2 " silk " " 1.00 " « • gloves " 1.00 " " 1 piece cambric muslin, 12 yds. at 1.00 " per yd. $20.41 Baltimore, June 11, 1840. Mr. Andrew Simpson, Bought of Robt. Broom, 3 bbls. superfine flour at $6.75 per bbl. 3 bushels of corn " 85 per bush. 8 " oata « 33 " $25.44 52 MULTIPLICATION. Baltimore, June 10, 1840. Mr. Wm. James Donohue, Bo't of Thos. Daugherty, 1 piece Irish Linen, 25 yds. at $1.12 1 " Bandanna hdkfs. " 10.50 1 " shirting muslin, 35 yds. " 37 6 " calico, each 29 yds. " 29 4 " Russia sheeting, cont. 120 yds. 45 $155.91 Baltimore, June 10, 1840. Bought of Thos. Greaves, Mr. Wm. Patridge, 28 yds. of broad cloth 15 bbls. of flour 90 gals, of molasses 14 lbs of coffee at u it M $5.60 per yd. 6.70 per bbl. 46 per gal. 15 per lb. $300.80 Mr. John Cantwell, 7 lbs. coffee 9 « tea 50 " fish 7 gallons of wine 1 bbl. oil Mr. Thomas Bruff, 16 lbs. coffee Baltimore, June 10, 1840. Bo't of Francis Jordan, at 14 cts. per lb. " 38 " " u 5 tt « " 1.25 " per gal. ¥ 23.82 " $39.47 Baltimore, June 11, 1840. Bought of George Conway, at per lb. 8 15 24 sugar hyson tea loaf sugar Mr. Wm. T. Beeks, 14 yds. tow cloth 40 " brown linen 4 pieces of nankeen 9 yds. striped jean 25 cts. « 42 " " « 17 a a $15.26 Baltimore, June 11, 1840. Bought of James Mooney, 8 cts. per yard. 25 " " 1.87 " " 20 " " $20.40 at a MULTIPLICATION. 5 Canton, July 1.1, 1840. Mi. David Johnson, Bought of Sam'l Smith, 1 8 lbs. tea at 98 cts. per lb. It. " coffee ' < 15 " M 36 " sugar * < 13 " M 1 7 M cheese c 9 « u 12 ;i pepper l < 19 " a 7 " ginger u 17 " U 1 3 " chocolate " 61 « ic $35.45 FARMERS 1 ACCOUNTS. I) . William Wallace, Jr. To Norman Nash, To 1 1 bbls. cider at $3.00 " 14 lbs. butter u 33 *• 52 « dried beef U 10 " 83 " cheese t< 9 44 4 bushels of apples a 25 " 3 firkins of butter, each 1 151bs. at 17 $109.94 LECTURE IV. ON SUBTRACTION. ( Sign - ) It is a fact well known to Mathematicians, that the piinciples of Arithmetic and Algebra, admit of in- crease, decrease, and equalitt. Increase consists of Addition and Multiplication; decrease, of Subtraction and Division. Equality, (being the result or answer agreeably to the data of the question, under different names,) is represented by a statement or equation, reduced bj increase or decrease, to the lowest term or answer. Hence, to facilitate increase or decrease, nothing remains but to simplify expressions or statements. After having learned to compose a number by the addition and multi- plication of several others, agreeably to the laws of in- crease, the first question that presents itself is, how to take one number from another that is greater; or, in other words, to separate the greater number into two parts, one of which shall be the given number. This is called the doctrine of decrease, and may be illustrated thus: Suppose we wish to take $3 from $10, by so doing we separate $10 into two parts, one of which shall be $3 the given number; then we begin with 10, the greater number in question, and descend as many places from 10 as there are units in the lesser number, and we shall come to the number required, which is $7. Hence, we find that 7 is the excess of 10 above 3, or we might SUBTRAC 1 53 say, that 7 is the (/ifference or remainder between 10 and I 'onsequently, the words excess, difference, and re- mainder, are synonymous, each answering to the separa- tion of $10 into the parts, $3 and $7, which is always designated by the name Subtraction. DEFINITION— TO SUBTRACT IS TO MAKE LESS. (Sign - ) Q. What is the upper line denoting the greater number calk A. The Minuend. Q. What is the lower line denoting the lesser number called? A. The Subtrahend. The difference of both is called the remainder. RULE. 1. Place the lesser number under the greater, so that units may appear under units, tens under tens, and hun- dreds under hundreds, &c. and draw a line underneath. Begin at the right hand, and take each figure in the lower line from the figure above it, and set down the re- mainder. 3. If the lower figure is greater than that above it, add 10 to the upper number, from which number, so increased, take the lower, and set down the remainder, carrying one to the next lower number, with which proceed as before, and so on, till the whole is finished. PROOF. Add the remainder to the lesser number, and if the sum be equal to the greater, the work is right. 56 SUBTRACTION. SUBTRACTION TABLE. 1 /\r«< f*esson. I From 1 Take 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 1 10 1 Rem. 1 o 1 1 i «l S| 4| 5| 6 1 7| Second lesson. Proa Take 2 •J 3 2 4 2 1 2 6 1 7 2 8 2 9 2 10 2 11 2 Ran 1 o 1 1 1 * 3| 4 | 6 | 6 | 7| 8| 9 Third Lnmn. From Take 2 2 4 3 6 3 61 7 :i | 3 8 3 1 10 3 11 3 12 3 Rem. o 1 1 | £| 3| 4|-6| 6 1 7| 1 Fourth Lesson. From Take •1 4 5 4 si a: 9 4 10 4 11 4 If 13 4 Rem. o 1 2| 3| 4| 5| 6| 7 | 8 | 9 1 1 . F\fth Lesson. , Frotn l Take 5 4 6 4 7 4 s 4 9 4 10 4 4 12 4 13 4 14 4 Rem. 1 * S| 4| 6| 6| 7 | 8 | 9 | 10 Sixth Lesson. From Take 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 Rem. M 2 3| 4| 5| 6| 8 j 9 | 10 Seventh Lesson. From Take 7 6 8 6 g i io ; ii 6 1 6 1 6 in 6 13 6 14 6 15 6 16 6 Rem. | 1 2 3| 4| 5| 6| 7 8 | 9 | 10 Eighth Lesson. From Take 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 Rem. | 1 1 _ 2 I .JJLlLLLLLl 8 | 9 j 10 Ninth Lesson. t From 1 Take 9 8 1\ 11 8 12 8 13 8 14 8 15 8 16 1 17 | 18 8 j 8 j 8 Rem. | 1 1 2| 3| 4| 5| 6 | 7| 8| 9 | 10 Tenth Lesson. Frotn | Take | 10 | •1 11 9 12 | 13 | 14 | 15 | 16 | 9 | 9 | 9 | 9 | 9 | 17 | 18 j 19 9 | 9 | 9 Rem. | M 2( 3 | 4'F 5 1 6 | 7 | 8 | 9 | 10 L _ — _ _ — . F SUBTRACTION. 57 1. Henry has 25 marbles and James 9, how many has Henry more than James? Ans. 16. 2. Thomas bought 10 oranges and gave 3 to William, how many had he left? Ans. 7. 3. Bought a book for 35 cents, and sold it for 25 cents, how much did I lose? Ans. 10c. 4. Bought a penknife for 75 cents and sold it for $1, how much did I gain? Ans. 25c. 5. If you give 25 cents for a knife and 20 cents for a book, how many cents did the knife cost more than the book? Ans. 5c. 6. The Arabian method of Notation was discovered in England about the year 1150, how long is it since to this period. Ans. 690. 7. Lucas De Burgo, discovered the method of book- keeping, by Double Entry, at Venice in 1495, how long since to this present year, 1840? Ans. 345. 8. If my son Ferdinand Augustine, had lived, he would be 3 years old on the 28th day of August, 1839, what time would he be 21 years old? Ans. August 28, 1857. 9. From $100 borrowed take $72 paid. 'Twas a lady lent it, what's due to the maid? Ans. $28.00. 10. From 90 take 30, from 40 take 10. Subtract 6 from 100, and what remains then? Ans. 184. SUBTRACTION OF DOLLARS AND CENTS. 1. From $166.94 From $9474.94 Take 149.68 Take 6496 69 Rtm'r. Rem'r. From $9476.97 Take 8469 69 From $6944.69 Take 4649.79 Rem'r. Rem'r. M I /TN xT "i\ / 4 4 \ V 1 1 / V 4 4/ The above cuts represent fractional parts of a cent and may serve to illustrate Subtraction, as well as Addition. inasmuch , as the latter can be proved by subtraction. From Take Proof. Add EXAMPLE. $978 73 i 644.37i $334.36* $644.37* 334.36i Here \ taken from leaves |. Total, $978.73 \ as above. This sum denoting the upper line, is called the min- ' vend, or greater number. 1. From $100 take * of a cent? Ans. $99,991 2. From $200 take 12* cents? Ans. $199.87$ 3. From $99.93| cents take $63.18$] Ans. $36,751 1. From $200 take 18| cents? Ans.$199.8U 5. From $130.37* cts. take $31,181? Ans. $99,181 6. From $199,111 cts take 931 cts.? Ans. $198.18 SUBTRACTION. 59 Example.— From $137.98c.7m. take $13.94c.8m. Ans. $124.03.9 n From $137.98.7 ' ,lium > Take * 13.94.8 Ans. $124.03.9 In the above example, composed of dollars, cents, and mills, the operation is performed the same as whole num- bers, except the pointing. Q. How is the pointing performed? A. By commencing at the units, and pointing off one place for mills, and two for cents. The reason is obvious, because there are 1000 mills in a dollar, which is equiv- alent to 100 cents. 1. From $193.98.5 take $37.63.9. Ans. $156.34.(> J. From $349.87.5 take $5.62.9. Ans. $344.24.6 3. From $869.93.9 take $8.67.9. Ans. $861.26 4. From $1000.08.9 take $9.68.9. Ans. $990.40 QUESTIONS FOR THE BLACK BOARD. 1. Subtract 37J cents from $10.00? Ans. $9, 2. Subtract 43 1 cents from $100.00? 3. Subtract 18| cents from $5.00? 4. Subtract $1.06 J from $18.00? 5. Subtract $3.37J from $6.93|? 6. Subtract $3.31 \ from $10.00? 7. Subtract $1.12* from $74.00? 8. Subtract $1.56i from $100.00? 1. From 139 Take 64 Remainder 2. M 86 << 49 it 37 3. « 128 u 99 M 29 4. " 402 it 315 u 87 5. « 616 a III C( 472 G. " 156 << 129 tt 27 7. " 124 tt 56 N 68 8. « 96 u 37 M 59 9. « 214 (( 177 (« 37 60 SUBTRACTION. QUESTIONS FOR EXERCISE. ADDITION AND SUBTRACTION. 1 . Sent a servant to market with $2 to buy provisions, who bought meat for 60 cents, butter 32 cents, cheese 28 cents, and fowls 30 cents, how much change must he re- turn? Ans. 50c. 2. Gave 25 cents for paper, 12$ cents for quills, and 37$ cents for an arithmetic, what change must I have out of a dollar note? Ans. 25c. 3. Borrowed $25.00 and paid $17.00, how much do I oweT Ans. $8.00 MULTIPLICATION AND SUBTRACTION. 1 . Bought the following items, viz: 16 lbs. sugar at 9 cts. per lb. 10 " cheese " 14 " " 4 " butter " 25 " " 6 " coffee " 13 " « $4.62 2. I handed the merchant a five dollar bill, how mud is the ^changeV Ans. 38c. 3. Bought 2 yds. muslin at 20 cts. per yard, 6 " calico " 25 " " 3 " ribband " 16 " " 3 cotton balls " 6 « 8 yds. green silk " 80 " «< $8.96 4. A merchant received a ten dollar note on the Ma- rine Bank in payment of the above bill, how much is the 'changeV Ans. $1.04 LECTURE V. ON DIVISION. ( Sign -*- ) Division consists of Addition, Subtraction, and Mul- tiplication^ in its operation, and in finding one of the fac- tors of a given product, when the other is known; like Subtraction, it is also, founded upon the principles of Decrease, and can be illustrated thus: If it be required to ascertain the number of times, 24 contains 4. We need only subtract 4 from 24, as many times as it can be done; and since, after 6 subtractions nothing is left, we con- clude 4 is contained in 24, 6 times; this manner of de- composing one number by another, in order to know how many times the last is contained in the first, is called Division; because, it serves to divide, or portion out, a given number into equal parts of which the number or value is given. The number to be divided is called the dividend, the factor that is known, and by which we must divide, is called the divisor, the factor found by the divi- sion is called the quotient, and always shows how many times the divisor is contained in the dividend. Hence, it is evident, that the divisor, multiplied by the quotient, will reproduce the dividend. 62 DIVISION. DIVISION OF THE NINE DIGITS ILLUSTRATED. 1 2 into 4 5 6 7 8 9 10 11 Quot'nt 2 2* 3 n 4 4i 5 5i 3 into 1 2 3 4 5 i 7 8 Quot'nt Oi Oi 1 H H 2 2i 2| 4 into 1 2 3 4 5 6 7 8 Quot'nt Oi Oi 0| 1 U H 11 2 5 into 1 2 3 4 5 6 7 8 Quot'nt Of Of Of Of 1 H If If 6 into 1 2 3 4 5 6 7 8 Quot'nt Oi Of Of Of Of 1 1J H 7 into 1 2 3 4 5 6 7 8 Quot'nt Of Of Of Of Of Of 1 If 8 into 1 2 3 4 5 6 7 8 Quot'nt OJ Of Of Oi Of Of 05 1 9 into 1 2 3 4 5 6 7 8 Quot'nt Oi Of Of Of Of Of 1 Of 10 into 1 2 3 4 5 6 7 8 Quot'nt Oy 1 , OA T s i rO T <, r oa oa o T v o T v _. (£y» The learner should be taught to set the remainders in a fractional form, and to know that a remainder is so many parts of the divisor, thus divide 6 by 7, and it leaves a remainder of 6, the remainder is 6 parts of 7, and is written £. Q. When you divide by 2, and have 1 of a remainder, what is the remainder called] Ans. i Divide by 3, and have 1, of a remainder. Ans. i Divide by 3, and have 2. Ans. § Divide by 4, and have 1, and so on. Ans. f Q. When we divide an apple into eight equal parts, what will each part be called? A. I into 9, into 10, into 11, into 12. DIVISION. 63 Q. What is the number given to divide by called? A. The divisor or denominator of a fraction. Q. What is the number to be divided by called? A. The dividend or numerator of a fraction. Q. What is that which is sometimes left after dividing, or when the operation is performed called? A. The remainder. Q. What is the number of times that the divisor is con- tained in the dividend called? A. The quotient. The dividend is a multiple of either the divisor or quotient, because the quotient x by the divisor is equal to the dividend. Q. If you have a remainder, must that remainder be less than the divisor? A. It must, or else there is an error. Q. If your dividend be dollars, what will your re- mainder be? A. Dollars. Q. If ounces, if cents, if pounds, if yards, &c? If years, if months, if weeks, if days, if hours, if minutes? What does the remainder shew? A. So many fractional parts of the divisor, and it must be of the same name with the dividend. Q. Divide $5 by 4, the quotient will be $1, and the re- mainder $1, what is that remainder called? A. i of a dollar. Divide $8 by 7, the quotient will be 1, and the remainder 1, which is f of a dollar, or ot the divisor. Q Can you *£-o' higher than 9 times in division? A. You cannot. Q. Why so? A. Because we have but 9 figures with a cipher (as before mentioned in addition,) in our numerical calcula- tions, as 1, 2, 3, 4-, 5, 6,7, 8,9,0. Consequently, the figure 9, is the highest number, which could possibly be placed in the quotient. 64 DIVISION. Q. How many parts in division? A. Three, the divisor, dividend, and quotient. Q. How is division proved? A. It may be proved by casting out the 9's as observed in addition, or by multiplication ; hence, multiplication and division mutually prove each other. Q. In what way? A. Because the rectangle or product of the divisor must evidently be equal to the dividend, Q. Can multiplication be proved by division? A. It can, by dividing the rectangle or product of the multiplier and multiplicand, by either as 16 X 12= 192 product. Here, 192 — 16, or 192 — 12. ~n "~i6 EXAMPLE. i I I ill 111 i i ^ 5 q or o o or 2) 6 (3 3) 6 (2 6 6 Because 2 is contained in 6 (3 times, and 3 is contained in 6) 2 manifest that 3 times 2 are 6, and that 2 ce 3 are 6. Again: •o . ~i ^ ill > :e § o a <y o o or 4) 8 (2 2) 8 (4 8 8 Here 4 is the divisor, 8 the dividend, and 2 the quo- tient, which shews that 4 is contained in 8, (2 times in like manner 2 is contained in 8 four times. Therefore, by reference to the multiplication table, we find that 2 ce 4 are 8, and 4 times 2 are 8. MMSION. 6i DIRECTIONS FOR TEACHERS. As soon as the pupil can recite the division table thoroughly, agreeably to the plan here laid down, let him be exercised in mental operations. Again, the next step will be to find the factors of given numbers, thus: 6 is a r of 24, how is the other found? Ans. from the table we know that 6x4 = 24, consequently, 4 is the other factor, and so on. l)i visor. Dividend . Quotient. I'roved. Thus nto 9 (3 times, because 3 times 3 are 9 « S) nto 12 ( 1 titn- I times 3 are 12 « 3)1 nto 15 ( 5 times, because 5 times 3 are 15 '• 3) nto 18 (ti timet, t> .■•<• iom ♦> tun''- :? in 18 « 3) nto 11 (7 times, because 7 times 3 are 21 " 8) nto 24 (8 times, because 8 times 3 are 24 M 3) Dt0 >~ (9 times, because 9 times 3 are 27 t< 4)i nto 12 (3 times, because 3 times 4 are 12 M 4). nto 16 (4 times, because 4 times 4 are 16 " < nto 20 (5 times, because 5 times 4 are 20 it 4) nto 24 (6 times, because 6 times 4 are 24 4) nto 28 (7 times, because 7 times 4 are 28 " 1 nto 32 (8 times, because 8 times 4 are 32 (« 4) nto 36 (9 times, because 9 times 4 are 36 :.) irtf o ( 1, because 1 or once 5 are 5 M 5) nto 10 (2, because 2 or twice 5 are 1<> M *) nto 15 (3 times, because 3 times 5 are 15 M 6) nto -jo (4 times, because 4 times 5 are 20 >. 5) irto IQ (5 times, because 5 times 5 are 25 " 5) DtO 80 (6 times, because 6 times 5 are 30 « 5) nto M (7 times, because 7 times 5 are 35 M 6) nt«. 40 (8 times, because 8 times 5 are 40 « 5) nto |0 (9 times, because 6 times 5 are 45 « 6) nto 6 ( 1 or once, " 1 or once 6 are 6 H 6) nto 12 (2 or twice, « 2 <n twice 6 are 12 « 6) nto 18 (3 times, because 3 times 6 are 18 « 6) nto 24 (4 times, because A times 6 are 24 « 6) into M (5 times, because 5 times 6 are 30 " «) lit.. ::»; (6 times, because 6 times 6 are 36 8* 66 DIVISION. DIRECTIONS FOR TEACHERS CONTINUED. Divisor. Dividend. Quotient. Proved. Thus: 6) into 42 (7 times, because 7 times 6 are 42 " 6) into 48 (8 times, because 8 times 6 are 48 " 6) into 64 (9 times, because 9 times 6 are 54 " 7) into 7 (lor once, " 1 or once 7 are 7 " 7) into 14 (2 or twice, " 2 or twice 7 are 14 " 7) into 21 (8 times, because 8 times 7 are 21 44 7) into 28 (4 times, because 4 times 7 are 28 " 7) into 85 (5 times, because 5 times 7 are 35 " 7) into 42 (6 times, because 6 times 7 are 42 " 7) into 49 (7 times, because 7 times 7 are 49 44 7) into 68 (8 times, because 8 times 7 are 56 44 7) into 68 (9 times, because 9 times 7 are 63 44 8) into 8 (lor once, *' lor once 8 are 8 " 8) into 16 (2 or twice, " 2 or 2 ce 8 are 16 44 8) into 24 (3 times, because 3 times 8 are 24 44 8) into 32 (4 times, because 4 times 8 are 32 44 8) into 40 (5 times, because 5 times 8 are 40 " 8) into 48 (6 times, because 6 times 8 are 48 " 8) into 66 (7 times, because 7 times 8 are 56 44 8) into 64 (8 times, because 8 times 8 are 64 44 8) into 72 (9 times, because 9 times 8 are 72 44 9) into 9 (1 or once, " 9 times 1 are 9 44 9) into 18 (2 or twice, " 9 times 2 are 18 44 9) into 27 (3 times, because 9 times 3 are 27 44 9) into 36 (4 times, because 9 times 4 are 36 44 9) into 45 (5 times, because 9 times 5 are 45 44 9) into 54 (6 times, because 9 times 6 are 54 H 9) into 63 (7 times, because 9 times 7 are 63 44 9) into 72 (8 times, because 9 times 8 are 72 44 9) into 81 (9 times, because 9 times 9 are 81 i C3r Let young scholars be lectured regularly, agreeably to the above plan, in order to be perfect in mental opera- tions. DIVISION. 67 SHORT DIVISION. RULE. 1 . Place the divisor to the left of the number you wish to divide. 2. Consider how many times the number by which you divide is contained in the first figure or figures of the number to be divided, and set down the result, noting if there be a remainder. 3. If there be no remainder, consider how often the divisor is contained in the next figure, but if there be a remainder, conceive it to be placed to the left of the next figure, into which divide as before and set down the result. APPLICATION OF THE RULE. Ex. 1. Ex. 2. Ex. 3. Ex. 4. Ex. 5. 2)482 2)648 3)963 4)484 2)236 241 324 321 121 118 Here in the 5th example, when we divide 2 into 2, the quotient is 1, which set down, but dividing 2 into 3 the next figure, the quotient is 1 and 1 over, which place be- fore the following figure 6, and 16 is represented; there- fore, 2 into 16 is contained 8 times. Hence, the quotient is 118 as above. Ex.6. Ex.7. Ex.8. Ex.9. 4)484 5)750 6)756976974 7)876942875 121 150 10. Ex. 11. 8)96869875 9)867487549874 The teacher is requested to give the pupil as many questions of this kind as he may think proper. 68 DIVISION. PRACTICAL QUESTIONS. 1. Two boys, Dick and Harry, have 12 apples which ^hey divide equally, how many will each have? Ans. 6 2. What is the quotient of 8736, divided by 8 and by 4? Ans. 273 3. Thoma6 gave 72 cents for 8 quarts of cherries, what was the price per q 1 . If four boys, John, James, William, and Robert, get 28 apples, how many will each have? 5. If Richard buys a slate for 6 cents, how many can he buy for 30 cents? 6. If a man travel 3 miles in an hour, how many hours will it take him to travel 36 miles? 7. At 6 cents a piece for oranges, how many can you buy for 48 cents? MENTAL EXERCISES. Q. How much is the half of $5.00. A. The half of 5 is 2 and 1 over, which is one dollar, and the half of a dollar is £0 cents; therefore, the half of $5.00 must be two dollars and fifty cents. Q. How much is the half of $7.00? Q. How much is the half of $11.00? Q. How much is the third of $1.00? Ans. 33 \ c. Operation, 3)1.00 33 1 cts. (£/» It is well known to the youth of our country that 1 00 cents is a dollar, and to the weakest capacity, that if an apple, a number or any thing be divided by 3, the -quotient will be $ of the dividend, therefore, 33 \ X 3 = 100. In the above case, it will be found that 3 would be a constant divisor, were ciphers to be annexed to the re- mainder ad infinitum, but by the principles of division when the divisors are constant, the quotients will vary as DIVISION. the dividends vary, the quotients, therefore, resulting from finite quantities divided by will vary as those quantities vary, and as finite quantities may be infinitely varied, an infinity of infinites is proven. Because it is well known to Mathematician*, that any finite quantity whatever, di- vided by will give infinites for a quotient, Q. — E. — D. Divide 2687676 by 2. Ans. 1343838 Divide 8469674 by 3. Divide 967485 by 4. 764756 by 5. 997349 by 6. 874966 by 934365 by 8. « 11 6796 1 77634 by 9. " 8626 Divide Divide Divide Divide Divide « 28232241 « 241 87 U " 152951$ « 166224| " 1249954 DI VISION OF FEDERAL MONEY. Divide $40.75 by 2. Divide 39.31 by 3. Divide 41.37 by 4. Divide 48.76 by 5. Diviee 37.45 by 6. Divide 44.18 by 7. Divide 50.49 by 8. Divide 100.00 by 9. Divide 50.61 by 10. Divide 55.67 by 11. Divide 60.73 by 12. Q. How do you write down one-half? A. J — I write 1, draw a line under it, and place 2 below the line, what is the figure 1 called? Ans. the numerator. Q. What is the figure 2 called? A. The denominator. Ans. $20.37$ " 13.10 " 10.34J 4 < 9.751 « 6.24| " 6.31+ " 6.3H « 11.11* " 5.06^ " 5.06 TT . " 5 06 T ' ¥ <t 4 H 18 H 6 a 26 1 8 H 46 u 9 u 37 a 10 u 44 u 11 a 23 H 11 M 68 70 DIVISION. EXERCISE QUESTIONS. Divide 2 into 14 Ans. 7 times, and no remainder. Remainder 2 a o « 6 « 1 H 4 « 1 u CASE II. When we divide by a composite number, say 12, U< lti, 18, 21,36, 43, 63, &c. de 144 by 12 — 4 times 3 are 12, now 4(144 3)36 12 Ans. Here we divide by 4 and by 3, because 4 X 3 =■ 12. Example 2. — Suppose we want to divide 196 by 14; here 2x7=14. Hence, we divide 196 by 2 and by 7, and the result will be the same as if divided by 14. Thus: 7)196 2)28 14 Answer. EXERCISES FOR THE SLATE. Divide II! by 16 m 160 « 40 " 256 < 4 64 " 567 " 24 « 625 « 25 « 1728 « 144 " 2744 " 196 DIVISION. 71 DIVISION AND SUBTRACTION. 1 . A gentleman made a present of $224 to his three daughters, in the following manner to the eldest he gave $80, and to the other two girls each, one-half, how much did each get? A. The oldest $80, and each of the other girls $72. J. A captain, mate, and 25 sailors, received a prize of $30,000, the captain got one-half, the mate one half what was left, and the remainder was to be divided equally be- lors, how much did each seamen get? A. Captain, $15,000, mate, $7,500, and each of the sailors, $300. QUESTIONS IN WHICH THERE ARE CIPHERS AT THE RIGHT HAND OF THE DIVISOR. EXAMPLE. A gentlemen sold a farm for $4,540, at $10 per acre, how many acres did the farm contain? Ans. 454 Rule. — Cut off as many ciphers from the divisor as then are ciphers in the dividend. Divisor Dividend. Quotient. Thus: 1(0 )454(0 (454 Ans. 1 . A gentlemen had a number of men in his employ- ment, to whom he paid $1,895.00, and each man received . how many men were there? Ans. 758 2. If 350 bushels of corn cost $217, what is it per bushel? Ans. 62c. 3. If 1000 gallons of molasses cost $430, what is it per gallon? Ans. 43c. 4. One hundred and forty-four men have to pay equal shares of a debt which amounts to $14400, how much must each man advance to make up the sum? Ans. $100. 72 DIVISION. QUESTIONS IN WHICH THE DIVISOR IS A COMPOSITE NUMBER. Example. — A farmer sold 28 cows for $448, how much did he receive for each cow? Ans. $16 ILLUSTRATION. It will he readily perceived that 28 is a composite num- ber, and that its factors are 4 and 7, as appears from the multiplication table, which is also a table of factors. Here, 28 divided into factors will be equal to 7 x 4, and ln-nce, we find that $448 divided by 7, will give 64 for a quo- tient, and that quotient, divided by 4 will give 16, exact- ly the same as if we divided $448 by 28. Operation, 7)448 or 28)448(16 28 4)64 — 168 Answer, 16 168 So that from the mode of cane appears that 1 refers to 16; consequently, one cow cost $16. 1. A lady paid $17.15 cts. for a quantity of cambric at 49 cents a yard, how many yards did she buy? Divisors. The factors are 7 and 7, because 7)171 fi 7x7 = 49. 7)245 35 Answer. 2. A merchant bought a quantity of beef for which he paid $6,480, at $15 a barrel, how many barrels did he buy? Ans, 432. Five and three are the common factors for 15, because 5x3=15, consequently, we divide by 5)6480 3)1296 432 dollars. DIVISION. 73 QUESTIONS FOR THE SLATE. Drride 10464 by 24? Ans. 436 " 16704 by 36? « 464 « 3379-2 by 48? " 704 « 62496 by 56? "1116 « 70119 by 63? "1113 « 80064 by 72? <fc 1112 « 93660 by 84? "1115 Miser. KL.WF.O IS EXAMPLES, To exercise the learner in Addition, Subtraction, Multiplication and Division. 1 . A farmer sold 487 bushels of wheat at $2 a bushel, what is the amount? Ans. ${)", How long will it take a man to travel 264 miles if he should travel 24 miles a day? Ans. 11. 3. In the field of battle there were 268 soldiers in rank, and 118 in file, what number of soldiers were there in the Army? Ans. 31624. 4. A gentleman had four bags of money, the first bag contained $3,475 the second $6,950, the third $934, and the fourth $467, how many dollars were there in the bags? Ans. $11,826. 5. A man bought 4875 bushels of wheat, and sold 1899 bushels, how many bushels had he left? Ans, 29*3 6. A merchant bought 198 barrels of flour at $5.00 a barrel, how many dollars did he pay for the whole? Anf. $990. 7. The sum of $2,000 was equally divided among 25 men, how many dollars did each man receive? Ans. 80. 8. A gentleman had 24 houses, and received 75 dol- lars rent, per annum, for each, how much did all his rents amount t Ans. $1800. 9. A gentleman divided his farm containing 895 acres, equally among his five sons, how many acres did each receive for his portion? Ans. 179. 7 74 DIVISION. 10. The earth moves round the sun at the rate of 69000 miles in an hour, how many miles does it travel in 24 hours] Ans. 2875. 1 1 . A merchant bought a bill of goods to the amount of $545, and paid at the time of purchase $450, how many dollars remain unpaid? Ans. $9500 12. How many acres are there in a piece of land 39 acres in length, and 20 in breadth? Ans. 780. 13. Suppose a man to earn $35 a month, how many months will it take him to earn $490. Ans. 14. 14. If $1,024 be divided equally among 64 men, how many dollars will one man receive? Ans. $16.00 15. In a certain school there are 36 scholars, among whom 540 quills are to be equally divided, how many will 1 scholar receive? Ans. 15. 16. If $40 pay for an acre of land, how many acres can be bought for $680? Ans. 17. 1 7. A farmer planted 2072 trees in 14 equal rows, how many did he plant in a row? Ans. 148. 18. If 35 yards of cloth cost $315, what would be the cost of one yard? Ans. $9. 19. If $774 be divided equally among 18 sailors, how many dollars will each sailor rec< Ans. $43. 20. If a man travel 48 miles a day, in how many days will he perform a journey of 3264 miles? Ans. 68. 2 1 . If a man travel 24 miles a day, in how many days will he perform a journey of 81b miles? Ans. 34. 11. A drover received $1188 for 44 head of cattle, how much did he get for each? Ans. $27. 22. The wages of 36 men for 1 month, is $540, how much will each get? Ans. $15. 23. Suppose a regiment of 648 men have 9720 pounds of beef, how many lbs. are there for each man? Ans. 15. 24. If 35 barrels of flour cost $210, how much did 1 barrel cost? Ans. $6. DIVISION. 75 25. If a regiment of 512 men have 8192 pounds of beef, how many pounds for each man? Ans. 16. litre we find 64 X 8 — 512; hence, as 64 is a com-' posite number we get 8 X 8 X 8 = 512. Operation, 8)8192 8)1024 8) 128 Answer, 16 26. If 105 yards of broad cloth cost $735, how many dollars does 1 yard cost? Ans. $7. DIVISION AND MULTIPLICATION. 1. A piece of cloth 13 yards in length was sold at auc- tion for $130, what is the price of 5 yards? Ans. $50, ILLUSTRATION. By division we can ascertain the price of 1 yard, and l»y having the price of 1 yard, we can determine the price of 5 yards by multiplication. Operation, 13)130(10 1 yard cost $10 13 5 $50 Ans. Note to teachers. — Require the learner (as in the above example) to find the price of 1 yard by division, and by multiplication, to find the value of any given number of yard*. \ V yards cost $96, what is the price of 9 yards? Ans. $36. 3. If 9 yards cost $36, what will 24 yards come to? Ans. $96. 4. If 8 yds. cost $32, what will 5 yds. cost? *< 20. %. If 5 yds. cost $20, what will 8 yds. cost? « 8& 76 DITISION. MEM \l. OPERATIONS. 1. If five pounds of coffee cost 75 cents, what is the cost of 2 pounds? Ans. 30c. 2. If 8 pounds of sugar cost 72 cents, how much did 3 pounds cost? Ans. 27c. 3. If 8 yards domestic muslin cost 96 cents, how much did 7 yards cost? Ans. 84c. 4. If 7 yards cost 84- cents, how much did 5 yards cost? Ans. 01 5. If 5 yards of broad cloth cost $15, how much will 9 yards cost at the same rate? Ans. $ 2 6. If 9 yards cost $27, how much did 5 yards cost? Ans. $15. 7. If 5 pounds of butter cost $1, how much did 3 lbs. Otfti Ans. 60c. 8. If 3 bushels of corn cost $1.50, how much did 2 bushels cost at the same rate? Ans. $1. 1 \KRCISE QUESTIO 1. A merchant paid $11.25 for a quantity of lemons, ts. a piece, how many lemons did he buy? Ans. 225 2. A farmer sold a quantity of pork for $30, at 8 cents per lb., how many pounds did he sell? Ans. 375. 3. A farmer sold a quantity of cheese for $29.82, at 7 cts. per lb., how many lbs. did he sell? Ans. 426. 4. How many oranges can you buy for $44.48, if you pay 8 cents a piece? Ans. 556. 5. A drover bought 12 oxen for $636, how many dol- lars did he pay for each? Ans. 53. 6. A gentlemen had 65 men in his employment to whom he paid for wages due $7,735, how many dollars had each man for his portion? Operation, 65)7735( Here we find that the figure in the units place, in the divisor and dividend is 5, which is a token, for a common measure; hence, 13 x 5 = 1547 cancel the 5 from both sides, and it is reduced to 13)1547 = 119 quotient DIVISION. 77 LONG DIVISION. Division is a compendious method of subtraction, it teaches how often one Dumber is contained in another, number to be divided is called the dividend, the num- ber to divide by is called the divisor, and the number of - the dividend contains the divisor, the quotient. — times there is a remainder left after the division is lied. C^. How arc the terms placed? A. The dividend is the middle term. The divisor is placed on the left hand side, and the quotient on the right. Q. In what manner? A. By curved lines each side of the middle term or dividend, as to divide 12 by 4, the quotient is 3. Dollars and cents may justly be considered as whole numbers and decimals, and are written as such. example 1. Divisor. Dividend. Quotient. Divide C3- 37($9869.75($266.75 74 246 249 222 277 259 is.') 185 Utistrate the above example ascertain bv how often the left hand figure (£/» (3) of the divisor is contained in the left hand figure £/* (9) of the dividend; we say 3 into 9 (3 times, we then multiply 37 i 78 DIVISION. which produce 111, and find 3 times too much; as, 1 1 1, is more than 98. Again, we try 2 \ twice) and find 37 x 2 — 74, we then set down 74 under 98, and take the difference, this difference or remainder is placed underneath the line drawn between it and the subtrahend 74. This being done, we next bring down the figure 6, which is placed to the right of the remainder, and find the number 246 represented as * new dividend; as before, we try how often the left hand figure 3 of the divisor is contained in 24, the 2 left hand figures of the new dividend, and find by trying that 6 times will an- swer, we then place the figure 6, after the figure 2 in the quotient, and multiply 37 by 6, which produce 222, and place it under 246, draw a line under, and take their dif- ference. We again bring down the next figure 9 which is placed to the right of the remainder 24, and the num- v!49 is represented as another new dividend. Again, we try how often the left hand figure fc^3 is contained in 2 1 . the two left hand figures of the new dividend, and find by trying as usual, that 6 times will answer — 6 times 37 are 222, which subtracted from 249 leaves a remainder of the next figure 7 is brought down and placed to the right of the remainder 27, and the number 277 is repre- sented. We proceed in like manner until the operation is performed, and find the quotient to be 266.75, we then commence at the right hand or units place, and point out two figures, 75 for cents, because there are cents in the dividend, consequently, two figures must be pointed. 1 . A gentleman has a yearly income of $75,920, how many dollars is that a day, there being 365 days in a year? Ans. $208. Diviaor Diridend. Quotient. Operation, 365)75920(208 730 2920 2920 • DIVISION. 79 Or thus: the units place in the divisor is supplied by the figure 5, and the units place in the dividend, by a cipher (0;) consequently, 5 will be a Hoken? for a divisor or common measure. Here 73 X 5 = 365, which are factors. Now, 5)75920 Quotient by short 73)15184(208 division as before. i n; 584 584 Seventy-three, the divisor in the last operation is a vrime number, and cannot be decomposed into factors. Q. What is a prime number? A. A prime number is that which can be measured only by itself, or a unit, as 7, 11, 13, 19,5, 23, 31, xc. Q. What is a composite number. A. A composite number is equal to the product of its factors or component parts as 28 = 7 X 4, 24 = 8 X 3. PROPERTIES OF NUMBERS. It is evident from analysis, that much time and labor can be saved, in the operation of numbers, and, that ques- which admit of many figures, can be readily an- swered orally, with precision and accuracy, agreeably to the following general rules, which are certainly of great importance, in elucidating the principles of the science. RULE. 1 . Figure 2 will divide all even figures without a re- main 2. Figure 3 will divide any sum, when the addition of the numbers can be divided by 3. 3. Figure 4 will divide any number of figures when the two last figures on the right hand, or the two last on the left can be divided by it without a remainder. 80 ^ion. 4. Figure 5 will divide any number of figures, if the last figure on the right hand be a 5 or a nought or cipher, (as 0.) Tigure 6 is a factor in all even numbers, which have ken of 3, or in other words, in any even number of figures, whose sum can be divided by 6, as 24864, (their sum make 24,) 6, will be a token. 6. Figure 7 has a variety of tokens. Case 1. — For two and three figures in a number, as 21, . 105, 126, I IT. 168, 189. When the left hand figure or figures, are double those on the right as above, 7 will divide without a remainder. Case 2. — By adding or subtracting 7, 14, &c. to or from the left hand figures the token would be obtained, as 168 -f 7 = 175 or 189—7 — 182; in either case, 7 will invariably be a div Case 3. — When the left hand figure divided into the two right hand figures, produce 5 in the quotient, as 315, '45, (as*3into 15) 5 times, 5)25(5 times, 9)45(5 times, figure 7 will be a general divisor or token; or, if 14,21, &c. be subtracted from any number, 7 will be a token, as 434, minus 14, leaves 420, winch is divisible by 7. Case 4. — For four figures in a number. When a divi- sion of the two left hand figures into the two right hand figures give a quotient of 5, as 1155, 7 will be a token; or, if the division of the right hand figures into the two left hand figures, give a quotient of 3, as 6622. Again, if two digits inclose two ciphers, as 1001 to 9009, 7 will be a token. Case 5. — For five figures in a number. When one cipher is inclosed by two equal numbers, as 27027, 7 will be a token. £5» It is really a fact, that 7, 11, and 13, are factors in those numbers, which have the token of one or two ciphers inclosed by two equal numbers or digits as 1001. DIVISION. % SI ILLUSTRATE j>pose it were required to multiply 485 by 11,7, and o a continual product, the answer can be obtained correctly without any calculation, by placing the same number on the side of it, thus: 485485, the product as red. Case 6. — In any given number of figures the token of 7 may be easily discovered by a division of the figures, ose 12684, 31563, 42525, 331121, 6301155, by in- spection, you can instantly discover the token in the three hand, and two right hand figures of the first. In the ttd number you see the token in the three left hand figures, also in the two right hand figures. In the third number you can readily discover the token, and in the last number it can be seen in the three left hand, and four hand figures. Rule 7. — Figure 8 will divide any number without a remainder, when the three right hand figures are divisible by it, as 1267848, ft Rule 8. — When any given number of figures added I or 102105, make 9, 9 will be a token; thus, in the first number the aggregate is 9, which can be !ed by 9, and the second number in like manner. It must be observed, that every number divisible by 9, is also divisible by 3, but every number divisible by 3, is not divisible by 9. Rule 9. — Figure 10 is a factor in all numbers, of which the last figure is a cipher. I. — By the figure 7, we find in what numbers 11 ctor. Q. Has it a general token? A. It has in all numbers, of which the figures sub- tracted from the right or kit hand, will leave no remain- der, as 88( s: DIVISION. Illustration. — 8 minus 3 leaves 5, and 6 minus 1 =■ 5, hence, 5 — 5 = 0or8— -3 = 5, and 6 — -5 — 1, and 1 — 1=0. '' . — If a figure cannot be subtracted from the next, ftdd 1 1 to it, and proceed as before. Thus, as in the num- ber 3267, thus: 2+11 — (13— 3) — 10, in liken with the right hand figures, 6 + 11 = (17 — 7) — 10, i'> — 10 — 0. Q. E. I). 1 1— Figure 12 is a factor in all numbers which have the tokens of 3 and 4, as 3 x i = 1 2. Figure 13 is a factor when the left hand figure or figures divided into the right hand figures, will give a quotient of 4. Rule 12. — Figure 14 is a factor in all numbers which have the token of 7. Figure 15 is a factor in all num- bers which ha kens of 3 and 5. Ruh n —Figure 25 is a factor in all numbers of which the two right hand figures divide without a remainder, as 1625, 3450, 7875, or a number which has two ciphers received on the end by a multiplication of 4. Rule 11. — Figure 125 is a factor when the thiee right figures can be divided by it, as 250, 89375, or when three ciphers are received on the end by a multi- plication of 8. Multiply 48 J - . in the multiplicand we discover the token of 25 x 4 = 100, or thus: Dividing 4;4812 1-203 x 100 = 120300. The product of any number to be multiplied by 25 may be obtained thus: place two ciphers at the end of the given number as 481200, which being divided by 4 will give 120300 the answer. Again, divide 789467125 !>v 1 2'^ agreeably to Rule 14, 125 is a token. Here 125 X 8 = 1000, hence, multiply the given number by 8, rejecting three places, and we have the quotient thus: 789467125 Multiply 8 6315737(000 DIVISION. 83 From the following easy multipliers and divisors, the labor of one hour can be done in a 10, 1 1, 12, 13, 11, 1 :., 16, 17, 18,19,20,21,31,41,51,61,71, 81,91, 100, 101, 901, &c, 10000, 100001, 900001, &c. PRELIMINARY REM ARKS. If we multiply any number by 10, 100, 1000, &c, we place the number of ciphers to the product. If we mul- tiply by 11, we place the figures of the multiplicand one place to the right or left hand under it, and add for the product. Thus: Multiply 7683 by 11 7G83 84-513 by placing to the right, 7683 11 7683 7683 84513 common method by placing to the loft. If we multiply by 15, 14, &c., we multiply the multi- plicand by the 5 or 4, and place the product to the right, as the 5 and 4 stand on the right hand of the unit, and add for the product. SECOND METHOD. as, 7365 by 14 or, 7365 29460 14 common way. 103110 29460 7365 103110 Thus you see, that the first method is the shortest. If we multiply by 101, 102, 901, &c, we place the >roduct two places to the right or left hand, as the figure >f the multiplier stands, and add for the product 84- DIVISION. Thus: 3465 by 106 20790 367290 The multiplier is to the right of the unit Again: 34-65 by 701 multiplier to the left of the unit. 24255 2428965 product. \N W.YSIS OF NUMBERS. Suppose it were n » analyse the number to its prime factors by inspection, agreeably to Rule 8, 9 is a token or factor, from whence we obtain 3168 =» 9 x 352, these factors again analysed, we get 3 X 3 X 4 X 88, because, ii c discover the token of 4, agreeably to Rule 3,3x3x4x88; :i x 3 are prime numbers and cannot be analysed. We shall analyse 4 X 88, 4 if a token, 1 x 4 = 4)4 X 22, expunge 4 both sides and what remains but 22; in this we find that 2 is a token, and we have 2x11, the prime factors are J, 2, 11. I have heretofore remarked, that agreeably to this plan, a person could in one minute perform a calculation which would require one hour's work in the ordinary manner to do it correctly. The following multiplication of factors will test the fact: EXERCISE FOR THE SLATE. Multiply 73678964 by 1429, 77, 28, 26, 25 and 5, continually, and tell the product Ans. 737747687568892000. ILLUSTRATION BY ANALYSIS. Multipy 73678964 by 1429, 77, 28, 26, 25, 5. Analysing 77 to 7 x 11, 28 to 7 x 4, 26 to 13 x 2. Multiplying 25 by 4, we have 2 ciphers, and 5 by 2, we get 1 cipher more. The operation stands thus: 73678964 by 1429, 7 x 11,4 X 7, 2 x 11, 100, 10. DIVISION. 86 By multiplying 1429 X 7 = 10003, in this number we find the tokens of 7, 11, and 1 3, and the work thus multiplied, we find the factors 7, 11, 13, remaining, which are tokens of 1001 product. 736789H1 by 10003 J J 1036892 737010676892 X 1001 737010676892 7M7717G87568892000 Answer. .ictors 10003 Multiply 1001 10013063 Product. Here by analysis the whole multiplication may be per- formed by a single multiplication of 3, almost in an in- stant. Multiply 931 by 325, 11- and 7 continually, and find the new factors and product. Ans. The factors are 13, 7, 11, 10, product, 93193100. The above continual multiplication may be performed by inspection, and the calculation verified for its certainty, in an instant. ANALYSIS OF MULTIPLICATION AND DIVISION. I VMPLE. Iculate the amount of 29 yards of muslin at 22 cts. per yard. Ans. $6.38. ILLUSTRATION. From the multiplication table 29 may be analysed thiu: J x 10 = 20 + 9, that is, 2 tens and 9 over, and in like manner 22 = 2 X 10 = 20 + 2 = 2 tens and 2 over. ce, from the properties of Euclid. 4, Lib. 2. *If i straight line be divided into any two parts the square of irhole line is equal to the squares of the two parts, together with twice the rectangle contained by the parts." Thus, let A B be divided into two parts, 6 and 4, the square of the whole line 10, which is 100, is 8 86 division. equal to 6 squared plus 4 squared, together with twice the rectangle or product of 6 x 4 = 24- X 2 = 48; hence, 10* «, 6« + 4« + 2)6 x 4 = 100 = (36 + 16+48) = 100. dolls, dime*, cents. First, 2 + 9 + 9 Multiply 2 by 2, and call the 2 + 2 + 2 product dollars, viz: $4.00 Add 9 and 2 the sum is 11, which multiply by 2, the product is 22 dimes, or - - 2.20 Again, 9 cents multiplied by 2, the product is 18 Proof by the common method 29 x 22 =* $6.38 Calculate the amount of 87 yds. at 83 cts. « " 73 " 85 M « « 35 " 38 " « « 95 " 97 " — $239.71 CASE II— When the figures under the head of dollars are different. Example 1 . Find the amount of 35 yards at 87 cents, $24.00 dolls, dimes. 3x7 = 21 3 + 5 + 5 8x5 = 40 8 + 7 + 7 Sum, 61 dimes. 61 dimes is 6.10 Then 7 cents X by 5 cents, is - - 35 $30.45 2. Calculate the amount of 83 yds. at 63 cts. « f* 54 " 35 " a « 27 « 49 " « « 99 " 39 « « « 56 " 78 « — $166.71 3. If 24 yards of cloth cost $43,92, the price of 1 yard is required? Ans.«$1.83. 87 Q. If there be cents or decimals in the dividend, and none in the divisor, how do you proceed in pointing? A. Point out as many figures in the quotient as there are cents or decimals in the dividend. Q. If there be as many cents or decimals in the divi- sor, as there are in the dividend, how do you proceed? A. The quotient will be expressed in whole numbers in that case, there being no difference between the deci- mals in the divisor and dividend. Q. If there should be more decimal places in the divi- sor than in the dividend. For instance, three decimal places in the divisor, and two places in the dividend? A. One or more ciphers may be added to those in the dividend. Then, the difference between the decimal places in the divisor and dividend will show the number of de- cimal places in the quotient when pointed. DIVISION OF FEDERAL MONEY. Divid e 140 dollars 85 cts. by' 15 quotient $9.39 « 485 " cc 5 a 97.00 « 694.42c.2ro. 1238 u 56.90 it 156 « m 4 a 39.00 a 10368 « ti 36 a 288.00 a 2688 " tt 112 a 24.00 "101442075 « u 4025 a 25203.00 "403524396 " u 44 n 83372.00 QUESTIONS FOR THE SLATE. Divide $1984911.185 by $44.25 Ans. $44856.75 M 4640.18i « 15.00 a 309,34$ it 7550.00 " 125.00 a 60.40 it 58.14 " 38.00 ti 1.53 a 1947.22 " 44.25c. f)/H. a ! 1 .50 tt 1562.67 « 6444 a 24.25 a 120.99 " 22.20 a 5.45 tt 49561776.00 "5137.00 it 9648.00 tt 2748900.00 " 350.00 a 7854.00 a 3250000.00 " 520.00 a 62500.00 88 DIVISION. Q. When there are ciphers in the dividend and divisor, how is the operation abbreviated? A. By cutting off, or expunging from the end of each as many ciphers, as are contained in the one that has the least number. Q. Divide 2150596750 by l Ans. 1720477 1. In this case, f) is a token for a divisor, agreeably to question 11; consequently, 125 = 25 x 5, but 25 is a composite number, then, 5 X 5 are its factors, and we get 5x5x5 = 125. 5)2150595750 Having only 30 figures in the ope- ration, exclusive of the answer. 5)430119350 5)86023870 1 : J04774 Answer. By the common method of division it will be found that there are 52 figures in the operation, exhibiting a dif- ference of 22 figures!! So much for lengthy, protracted, old-fashioned methods. APPLICATION. 1. Bought 64 yards of cloth for $496, what was it per yard? Ans. $7.75. 2. Sold 27 bushels of wheat for $33.75, how much was it per bushel? Ans. $1.25. 3. If 13 pounds of butter cost $3.51, what was it per pound? Ans. 27c. 4. If 19 barrels of cider cost $42.75, what was it per barrel? Ans. $2.25. 5. If 35 hogsheads of molasses cost $551.25, what was it pr. gallon? (allowing 63 gals, in a hhd.) Ans. 25c. 6. Bought 3 pieces of domestic muslin, each piece containing 29 yards for $13.05, what was it per yard? Ans. 15c. 7. Bought 12 bushels of apples for $7.80, how much did 1 bushel cost? Ans. 65c. DIVISION". 89 ADDITION AND SUBTRACTION, 1. If I add 60, 90, and 560, and subtract from their amount 30, 40, and 90, what number will remain? Ans. 550. J. If in 20 chests, be 20 drawers, in every drawer be 20 purses, in every purse $20, how many purses and dollars do the 20 chests contain? Ans. 400 purses, and $8000. ADDITION AND MULTIPLICATION. 1. There are 10 bags of coffee, weighing each 160 lbs., and 12 bags weighing each 135 lbs., what is the weight of the whole? Ans. 3220 lbs. SUBTRACTION AND MULTIPLICATION. 1. There are 15 bags of coffee, each weighing 112 lbs., the bags which contain the coffee weighing 30 lbs., how much would the coffee weigh without the bags? Ans. 1650 1b*, LECTURE VI. ON DECIMALS. Whoever may have been the inventor of Decimals. Moses, the Lawgiver of the Jews, the Cotemporary of Cadmus, appears to have been among the number of those, who first employed them to any considerable ex- tent. Independent of any extraordinary divine influence upon the mind of this individual, he was as a man, one first, during the times in which he lived. " He was educated at the Court of Pharoah, "in all the learning of the Egyptians," which was, of course, in all the sciences then known. This learning was necessarily communi- cated in connexion with the Egyptian system of hierogly- phics; therefore, he was aware, of the difficulties his coun- trymen had to encounter, in relation to their land marks being defaced by the ebbing and flowing of the river Nile, and in consequence thereof, decreed, that each indi- vidual should provide for his own subsistence, without seizing what was in the possession of another; hence, necessity and self-preservation gave rise to the useful arts. It was then, the science of Geometry became known. It was then, that man beheld with new eyes the magnifi- cent spectacle, which nature exhibited to his senses and imagination. It was then, he learned to compare and examine; and it was then, his ideas were transported, as it were, into an intellectual world. He studied the phe- nomena of nature with discriminating attention, till his mind was impressed with a desire to know the causes by which they were produced. From the history of Jose- DECIMALS. 91 phus, we have been informed that the Patriarch Abraham, im Chaldea into Egypt, during the time of a famine, and taught the inhabitants of certain portions of that country a knowledge of arithmetic. Also, that his mode of instruction was based upon the method of No- ;i, substituting Hebrew letters for modern numerals, **Aleph 1, Beth 2, &c. to Yod 10, Caph 20, to Keph, 100, &c, reckoning singly from 1 to 10. So that it is evident, he understood the decimal scale or decuple divi- sion of numbers: these facts and their resulting influence are traced as far as human knowledge extends; in short, they are from the very time of the invention of letters. Henc< . it i- natural to suppose, that some knowledge of numbers more or less perfect, must have been coeval with human society, from the bare consideration of our natural wants and early impressions. mil: denary or decimal scale of tenths. Q. What is a decimal fraction? A. It is a fraction having always some power of 10, denominator. Q. When an integer or whole is supposed to be divided any given number, as T V, T iv> T^nr> &*•> how is it essed or written? A. As .2 .02 .002, &c. with a point (.) to the left. Q. What is the general rule in this case? A. To point off' as many figures as there are ciphers in the denominator; the places between the significant figure and the point being supplied with ciphers if necessary, as above. Consequently, the same number of figures on right of the decimal point, has always the same de- nominator. Q. How is the value of digits ascertained agreeably to the law of increase from the Denary Scale. 92 DECIMALS. A. From Notation we know the value of figures in- credses in a ten-fold proportion from the right to the left. 4 • Thus:- . ,, I 2 I . ? • i 1111 r= 1000+ 100 4- 10+ 1 = 10 + 10 + 10 4- 1; hence, it is evident, that the distance of any figure from the right indicates the power of 10. Q. How is the value of decimal figures ascertained agreeably to the law of decrease? A. In mixed numbers the decimals are separated from the integers by a point; thus: 25 T $ V is written 25.02, agreeably to question 4. "We point out as many deci- mal places as there are ciphers in the denominator." Consequently, it can be easily perceived that the value of als decreases in the same ten-fold proportion from the point (.) towards the right hand, as that of integers, increases from the point towards the left. In Geometry we find that lines emanate or flow from a point (.); so in Arithmetic, the power of numbers is derived from a unit, and although we can by the pre- ceding rules apply to fractions in all cases. The four fundamental operations of arithmetic, yet it must have been long since perceived, that if the different sub-divi- sions of a unit employed for measuring quantities smaller than the unit, had been subjected to a common law of decrease the calculus of fractions would have been much more convenient on account of the facility with which we might convert one into another. By making this law of decrease conform to the basis of our system of numer- ation, we have given to the calculus the greatest degree of simplicity of which it is capable. ILLUSTRATION. Now as a unit is conceived to be divided into 10 equal parts, each part would be a tenth, and if each of these I IMALS. 93 parts or tenths were again divided into 10 paits, each would be one-hundreth part of a unit. Again, if these were again divided, in like manner, each part would be one- thousanth part of a unit and so on. Example, 1 cent or T $ v part of a dollar is expressed thus: .01, agreeably to question 4; because, we must point out two places for cents invariably. NUMERATION OF DECIMALS. EXAMPLES. tftift* down a unit? 1 10 1 100 1 1000 $1.00 Unit. down decimally? M 0.10 Hundreths. 0.01 Hundreth. 0.001 Thousanth. 1 10000 « 0.0001 Tenths of thou- sands. << 1 " 0.00001 Hundreths of thou- sands. 100000 .. 1 M 0.000001 Millionth of a unit or dollar. 1000000 1 M 0.0000001 10 Millionth part of a dollar. 10000000 0.00000001 100 Millionth part part liar. 100000000 of a unit or do! ADDITION OF DECIMALS. ( Sign + ) Rule. — Add as in whole numbers. Add 65.43.9 Add 400.03.9 Add 4.00.30 " 84.: " 640.72.6 " 54.97.40 " 640.65.4 « 3.21.00 150.n:. « 947.67.6 « 6.72.03 68.90.73 N DECIMALS. Add 647.83.9 Add 43.00.7 Add 427.00.0 " 494.74.7 CI 64.73 8 " 603.04.0 « 648.93.5 a 79.89.4 " 210.15-0 " 943.49.7 M 38.49.6 « 3.36. " 846.79.9 U 44.48.4 « 0.02. " 469 94 6 TUi/n/Tiu 1243.57.0 MULTIPLICATION OF DECIMALS. (S 'gn X ) Rule. — Point off as many decim als in the product as are in both Factors. 1. Multiply 231. 11 :> by 8 Ans. 1851.320 J. " 32.1509 SI 15 if 482.2635 3. " .840 a 840 a 705.600 4. " 1.236 u 13 a 16.068 5. " 223.86 u 2.500 ti 559.6500 6. " 35.640 M 26.18 « 993.05520 7. " 8.4960 M 2.618 u 22.2425280 8. " 0.5236 « 0.2808 a 0.14702688 9. « 0.11785 U .27 u 0.03181950 CALCULATION OF BILLS BY DECIMALS. J. Thompson Laws, 1 ? yds. of flannel 19 "shalloon 16 u blue camlet 13 patterns silk vestings 9 yds. cambric muslin 25 " bombazine 17 " ticking 19 " striped jean Bought of John Jones, at 47 cts. u 37 « u 46 t< « 3.78 ft a 63 « « 56 a a 31 « a 16 u $99.50 DECIMAI 95 EXERCISES FOR THE SLATE. 1. What will 126 lbs. butter cost at 12c. pr. lb? $15.12 & What will b3 lbs. tea cost at 65c. per lb? 40.95 3. What will 13 tons of hay cost at $ 15.75 pr. ton? 204.75 4 . What will 46 lbs. pork cost at 7c. 5m. pr. lb? 3.45 What will 76 cwt. beef cost at $5.28 pr. cwt? 401.28 6. What will 19 cords of wood cost at $3.75 pr.c? 71.25 7. What wiH7b'9 lbs. of leather « .195 pr. lb? 149.95* 8. What will 65 gals, of wine « $1.75 pr. gal? 113.75 9. What will 796 lbs. of cocoa « $6.71 pr. lb? 5341.16 10. What will 19bbls.ofcider « $1.37$ pr.bbl. 26.12* 11. What will 49 green hides « $1.95 95.55 12. What will 15* gross bottles of castor oil cost at .375 per bottle? - - 810.00 13. What will all the above items amount to? $7273.34 CASE I. Rule 1. — If the number of figures in the product is not so great as the number of decimals in the multiplier and multiplicand, a sufficient number of ciphers must be placed to the left of the product, to make the figures in the product equal to the number of decimals in both fac- tors, and the decimal point must then be placed to the Ml of the ciphers, as in example 4th. EXAMPLE 1. 1. Multiply - - - .18 by 07 .07 Agreeably to Case 1, Rule 1, .0126 2. Multiply .008 by .08. 3. Multiply -0008 by .008. 4. Multiply $1.05c.3m. by one dollar and 2 cents. 5. Multiply .50 cents by .50 cents. Ans. 25c. 6. Multiply .10 « by .10 « " lc. • 144 or 12 dozen = gross. !>kCl.MAl.>. 7. Multiply .60 cents by .60 cents. Ans. 36c. 8. a 9. u 10. a 11. u 12. u 13. M 14. U 15. H 16. M 17. u •25 " by .25 .75 " by .75 « .40 ^ by .40 " 1 cent by 1 cent. 5 cents by 5 cts. A, .20 " by 20" .30 « by .30 " 6ic, « 56ic. " 16c. « .0001c. i of a ct. or .0025 Ans. 4c. Ans. 9c. .50 « by .75 « Ans. 375 or 37*c. .345 « by 6.25 cts. Ans. 2.15625 5 mills by 5 mills. « 0.000025 QUESTIONS FOR THE SLATE. Multiply .18 361 54.2 4560. .28043 .00071 4.001 by . 1 by 2.5 by 38.63 by .372 by .0005 by .121 by .004 Ans. 4.32 « 90.25 Ans. 2093.746 " 1696.320 Ans. .000140215 « .00008591 " .016004 SUBTRACTION OF DECIMALS. (Sign-) The rules prescribed for the subtraction of whole num- bers apply also to decimals. From 4562 From 86.746 From 978.4464 Take 3160 Take 73 638 Take 435.4969 1402 From 9648.6978 From 986974.668875 Take 4697.6986 Take 745649.954689 1. From $100, take 1 mill? Ans. $99.99c.9/n. 2. From $400, take 1 cent? Ans. $399.99 3. From $750, take $5. Op. 5? « $744.94.5 4. From $110, take $4,44 4? « $105.55.6 im its. 97 •.ike five dollar-. <i five mills? '. take $1, one cent and 9 mills? . take $1, nine cents and 3 mills' >m seventy-five dollars, seven cents and 7 mills, tine dollars, nine cents, and nine mills! DIVISION OF DECIMALS. ( Sign - ) Division of Decimals is performed the same as in whole numbers, only observing, that the number of deci- mal- in the quotient must be equal to the excess or differ- ence between the number of decimals of the dividend, and those of the divisor. When the divisor contains more decimals than the dividend, ciphers must be added or affixed to the right hand of the latter, to make the num- ber equal to, or exceed, that of the divisor. Example 1.— Divide 14.625 by 3.25. Divisor. Dividend. Quotient. 3.25)14.625(4.5 1300 1625 1625 Answer. In this example there are 2 decimals in the divisor, and 3 in the dividend, their difference 3 — 2 = 1; hence, only 1 decimal is to be pointed off in the quotient. Example 2.— Divide 3.1 by 0.0062. Q. Which is the divisor? A. Tin- number to the right of the word *£y' when set down, as, 0.0062. Q. Which is the dividend? A. The number to the left of the word %,' as 3.1 . Q. How it the operation performed? A. The same as in division of whole numbers. DECIMAL. Q. When there are moie decimals in the divisor than in the dividend, how do you pn A. By prefixing as many or more ciphers (as the case may be) to the right of the dividend as there are decimals in the divisor. Divisor. Diridend. Quotient. Illustration, .0062)3.100000(500.00 8.10 0000 Ans. From this operation, the learner can see there are 6 - of decimals in the dividend, and 4 in the divisor, and by taking 4 from 6, the remainder is 2, which shows the number of decimal figures to be pointed off in the quoti. CASE I. — When there are as many drcinv divi- sor as in the dividrnd. Q. How is the quotient expressed? A. In whole numbers. Give an example* Divide 9.6 by .06. Here, by prefixing a cipher to 9.6 it becomes 9.60, and then, has 2 decimals, the same as in the divisor 06)9.60( 1 60, consequently, 160, is a whole number. 1. Divide 17J236 by 1.16 Ans. 14.8750 2. « 148.630 by 1.21. " 35.304 + 3. « 21 12. by « 66.9375 4. " 2.00383 by 931. " 0.0021523 5. " 64.395 by 40.5 « 1.59 6. « 5.8674 by 127. " 0-0462 7. « 2033.100 by 0.324 « 6275.— 8. " 1383.2 by 60.8 " 22.75 APPLICATION OF DIVISION AND MULTIPLICATION, 1. Divide $1 by five cents? 2. Divide $5 by 5 mills? 3. Divide $44.44 by 11 mills? 4. Divide $44.50 by 5 mills? DEC1MV 99 of cloth cost $4.50, how much did 1 Ans. .37.5 or 37£c. a cost $1.25, how much did 3 yards Ans. .9375 or 93f c. 7. [f 5 yards cost .62.5, how much did 8 yards cost? Ans. $1 8. If 7 yards cost .87.5, how much did 2 yards cost ? Ans. .25 9. If 3 yards cost .37.5, how much did 7 yards cost? Ans. .875 10. If 7 yards cost .4375, how much did 16 yards cost? Ans. $1 11. If 16 yards cost $1, how much did 3 yards cost? Ans. .1875 U. If 5 yards cost .3125, how much did 8 yards cost? Ans. 50c. 13. If 3 yards cost .1875, how much did 9 yards cost? Ans. .5625 14. If 8 yards cost 50c. how much did 12 yards cost? Ans. 75c. REDUCTION OF DECIMALS. From what has been said, the following questions and a era are sufficiently evident, in relation to reduction of decimals. CASE I. I ) How is a number of a lower denomination changed e decimal of a higher? A We first suppose it to be changed into a fraction, having 10, or some multiple of 10 for its denominator. Q. How do you proceed next ? A. We add ciphers to the numerator, and divide it by so many as make one of this higher denomination, and the quotient is the required decimal. Q. Can the decimal thus obtained be again converted into a decimal? tOp DKCIMALS. A. Yes, agreeably to the decimal scale it has 10, or a multiple of 10 for its denominator, and by division it can be reduced to a .^till higher name. CASE II. — To reduce a decimal of a higher name to a lowi r. Q. How is this kind of reduction performed? A. We multiply the decimals by so many as make one of a lower denomination, and point off the decimals as already observed in multiplication. The figures which remain on the left of the period, when the proper number is separated for decimals will constitute, the whole num- ber of this denomination. Q. Can the decimal part be still reduced? A. Yes, if there be lower denominations, multiply it !>v the number which is equal to one of the next denomi- nation, and proceed as before. AVOIRDUPOIS WEIGHT. CASE I. — To bring a lower denomination to a higher. 1 . Reduce 8 cwt. to the decimal of a ton. ILLUSTRATION. The highest denomination mentioned is a ton, which is 20 cwt, then the fraction will be expressed thus: a or f , agreeably to question 8, annex two ciphers OQ to the numerator **° = .40 Ans. Bring 7 cwt. to the decimal of a ton? Ans. .35 3. Bring 5 cwt. to the decimal of a ton? * .25 4. Bring 14 lbs. to the decimal of a cwt? " .125 5. Bring 21 lbs. to the decimal of a cwt? Ans. .1875 6. Bring 3 quarters to the decimal of a cwt. " ;75 7. Bring 2 qrs. 14 lbs. to the " of a cwt. « .625 8. Bring 5 ounces to the decimal of a lb. " .3125 T>FJ 101 C AS E 1 1 . — To reduce a decimal of ah igher name to a lower. i . fa ton to its proper value? 20 8.0 Answer 8 cwt. >duce .35 of a ton to its proper value? Ans. 7 cwt. ■\. Redact .25 of a ton to its proper value? " 5 " 1. Reduce .125 of a cwt. to its proper value? " 14 lbs. 5. Reduce .1875 of a cwt to its proper value? " 21 " Reduce .75 of a cwt. to its proper value? " 3 qrs. 7. Reduce .625 of a cwt to its proper value? Ans. 2 qrs. 14 lbs. v Reduce .3125 of a lb. to its proper value? Ans. 5 oz. APOTHECARIES WEIGHT. CASE I. — Bringing a lower number to the decimal of a higher. 1. Bring 8 ounces to the decimal of a pound? 8 ss 3 reduce $ to a decimal add two Example 4 — ciphers to the numerator * J ° = 12 .66 -f. -J. Reduce 2 scruple* to the decimal of a pound? Ans. .00794 A. Reduce 18 grains to the decimal of an ounce? Ans. .0.0375 LAND MEASURE. CASE I. 1 . Reduce 20 perches to the decimal of an acre? Ans. 425 deduce 32 perches to the decimal of an acre? Ans. .20 3. Reduce 16 perches to the decimal of an acre? Ans. .0.10 4. Reduce 2 roods, 32 perches, to the decimal of an acre? Ans. .7 r 103 DECIMALS. Keduce 3 roods, 20 perches to the decimal of an acre? Ans. .875 »'. Reduce 3 roods, 35 perches to the decimal of an Ans. .96875 CA8E II. 1. Bring .125 decimal of an acre to its proper value? Ans. 20 perches. J. Bring .20 of an acre to its proper value? Ans. 32p. I. Bring .60 of an acre to its proper value? Ans. 2 roods, 16 perch 4. Bring .96875 of an acre to its proper value? Ans. 3 roods, 35 perches. Bring .4-5 of an acre to its proper value? Ans. 1 rood, 32 perches. •. Bring .40 of an acre to its proper value? Ans. 1 rood, 24 perches. CLOTH MEASURE. CASK I. 1 . Reduce 2 quarters and 2 nails to the decimal of a Ana. .625 2. Reduce 3 quarters and 3 nails to the decimal of a Ans. '.937") & Reduce 1 quarter and 1 nail to the decimal of a yard? Ans. .3125 CASE II. 1 . Bring .625 of a yard to its proper value? Ans. 2 grs. 2 nail-. J. Bring .9375 of a yard to its proper value? Ans. 3 qrs. 3 nails. ^. Iking .3125 of a yard to its proper value? Ans. 1 qr. 1 nail. k Bring .865 of a yard to its proper value? Ans. 3 qrs. 2 nails. Bring .375 of a yard to its proper value? Ans. 1 qr. 2 nails. 103 APPLICATION. I. What will 3.875 acres come to at $10 per acre? Ana. $38,750 What will 15 125 acres cost at $30.75 per acrer^ Ans. $465,093 What will 10.625 cwt. of iron cost at $5.25 pr. cwt? Ans. $55,781 4. What will 10.1875 cwt. " cost at $5.35 per cwt? Ans. $54,503 V hat will .I-:.") cwt. " cost at $7.10 per cwt? Ans. .887 6. What will .3125 lbs. of tea come to .87c.5m. per lb? Ans. .273 7. What will .35 of a ton come to at $96.50 per ton? Ans. 33.775 & What will .4 of a ton cost at $95 per ton? s Ans. $38.00 9. What will .625 of a yard cost at $4.50 per yard? Ans. $2,812 10. What will .875 of a yard cost at 37c. 5m. per yard? Ans. .328 I I . What will .8 of an English Ell cost at $3.25 pr. yd? Ans. $3.25 U. What will ,1875 of a yard cost at $5.00 per yard? Ans. .937 13. What will .375 of a ton cost at $80 pr. ton? « $30 ! 1. What will .859375 bushels come to at $1.3125 per Ans. $1,128 1 "». What will .875 lb. avoirdupois cost at 15c. per lb? Ans. .131 A hat will 3.75 gal. cost at 33c. pr. gal.? *« 1.237 17. What will .125 gals. .375 " « .047 18. What will 333.75 ft. of boards come to at 3c. pr. ft? Ans. $10,012 19. What will 33 T ' T dozen buttons cost at 87c. 5m. per dozen? Ans. $29,385 LECTURE VII. ON REDIVI l«'\. Reduction admits of Multiplication and Division, and of course, the signs x and -~ are understood, when in reference to these rules. Reduction shows, how we of names in use, May high to low and low to high reduce; So that the answer, which shall thence arise, The given sum in value equalize. Reduction, in other words, is an application of Multi- plication and Division: when higher quantities are reduced to lower, as years to seconds, Multiplication is used and the operation is called Reduction descending, illustration. pose we bring 1 year to months; we know agreea- bly to the law of increase, that the operation is performed by multiplication; hence, we multiply by 12, to bring years to months. On the contrary, division is used in re- ducing lower names to higher, as when we want to re- duce (lbs.) pounds to hundreds, (or cwts.) we divide the pounds by 28, to bring them to quarters, the quarters by 4. to bring them to cwts. This reduction is called reduc- tion ascending. FEDERAL MONEY, OR UNITED STATES CURRENCY. The denominations are — 10 mills (m) make 1 cent, - - c. 10 cents (cts.) " 1 dime, - - d. 10 dimes " 1 dollar, - $orD. 10 dollars " 1 Eagle, - - E. I REDUCTION. 105 AVOIRDUPOIS WEIGHT, v thing of a coarse drossy nature is bought and sold by this weight. FIRST FORM OF THE TABI.K. b 20 hundred weight make 1 ton, T. 4 quarters or 112 lbs. make 1 hundred, cwt. 28 pounds make 1 qr. of a cent, qr. 16 ounces " 1 pound, lb. 16 drams " 1 ounce, oz. Q. How are these numbers used? A. As multipliers, when a higher name is reduced to a lower. SECOND FORM. 16 drams make 1 ounce, oz. 16. ounces " 1 pound, lb. 28 pounds '* 1 qr. of a hun. qr. 4 quarters " 1 hundred, cwt. 20 hundred weight make 1 ton, T. Q. How are the numbers in the second form of the table of Avoirdupois weight used? A. As divisors, when bringing a lower name to a higher. CASE I. — Bringing a higher name to a lower, .. Example 1. Bring 2 tons to drams? Ans. 1146880 drs. Q. What are tons multiplied by? A. By 20. Q. Why? A. Because 20 hundreds make a ton. Q. What are cwts. multiplied by? A. By 4. Q. \\ A. Because 4 quarters make 1 cwt. weight. Q. What are quarters multiplied by? A. Q. v. A. Because 28 lbs. make a quarter. Q. What are pounds (lbs.) multiplied by? A. By 16. Q. Why? 106 REDUCTION. A. Because 16 ounces make a pound (lb.) What areounces multiplied by? A. By 16. Q. Why? A. Because 16 drams make an ounce, (oz.) Q. What does the first form of the table refer to? A. To reduction descending. Q. Recite rule Isti A. Tons multiplied by 20 are cwts. Cwts. " by 4 are quarters. Quarters " by 28 are lbs. Pounds " by 16 are ounces. Ounces by 16 are drams. The learner is earnestly requested to recite the table of Avoirdupois weight, agreeably to Rules 1 and 2, as laid. dowa in this work. tans. cwt. qrs. lbs. ozs. drair.8. \mple.— Bring 1 17 :i 17 11 13 to drams? 20 By Rule 1. 37 cwts. 4 151 quarters. 28 1215 303 4245 pounds 16 25474 4246 67934 16 407607 67935 10S6957 drams. 1<>7 ILLUSTRATION. we multiply the tons by 20, and add <tV the 17 cwts. and we get 37 cwts. The cwts. we multiply by 4, and add in 3 quarters, which make 151 quarters; the quarters are multiplied by 28, and 17 lbs. are added, which make 4245 lbs. The pounds are multiplied by 1<>; and 14 ounces are added, which make 67934 ounces. Again, the ounces are multiplied by 16, and 13 drams are added; hence, we get 1086957 drams for the answer re- turned. APPLICATION. I. Bring 13 tons to cwts? Ans. 260 cwts. Bring 260 cwts. to quarters? Ans. 1040 qrs. ::. Bring 36 quarters to pounds? " 1008 lbs. Iking 17 pounds to ounces? " 272 ozs. Bring 20 ounces to drams? " 320 drs. - duce 3 quarters, 14 lbs. 11 ozs. 11 drs. to drams? Ans. 25275 drams. '. Reduce 5 tons, 12 cwt. 2 qrs. to quarters? Ans. 450 qrs. 8. Reduce 2 qrs. 14 lbs. 10 ozs. to drs. « 18080 drs. 9. Reduce 12 cwt. 1 qr. 22 lbs. to lbs? " 1394 lbs. Illustration.— To analyse 112, we get 100 + 12 = 112. Common method. Oral Reduction of cwts. qrs. and lbs. cwta. qra. lbs. cwts. 12 i ■:■: 12 x ioo — 1200 4 12 x 12 = 144 — 1 qr. and 22 lbs. - 28 -f 22 =* 50 49 28 Answer, 1394 lbs. 100 I lbs. answer. 10. Bring 15 cwt 3 qrs. 20 lbs. to lbs? Ans. 1784 lbs. II. Bring 18 " 2 " 18" to lbs? " 2090 " 108 REDUCTJ 1 J. Bring 38 cwt. 3 qre. 20 lbs. to lbs? Ans. 4360 lbs. 13. Bring 64 * %« 14 « to lbs? " 634'J APOTHECARIES WEIGHT By this weight Apothecaries mix their medicines, they give only 12 ounces to a pound. nasT roan or THE TABLE. The denominations are — 12 ounces make ams " 1 pound, 1 ounce, ft 3 3 scruples " grains (gr.) make 1 dram, 1 scruple, 3 9 Q. How are these numbers used? A. As multipliers, when a higher name is reduced to a lower. SECOND FORM. 20 grains make 1 scruple, 9 3 scruples u 1 dram, 3 8 drams " 1 ounce, 3 12 ounces " 1 pound, Jfe Q. Recite Rule 1st? A. Pounds multiplied by 12 are ounces. Ounces " by 8 are drams. Drams M by 3 are scruples. Scruples " by 20 are grains. CASE I. — Bringing a higher name to a lower. Example. Rule 1st. — Bring 28 lbs. to ounces? L2 Answer, 386 ounces. 1. Bring 72 ounces to drams? Ans. 576 drs. J. Bring 10 lbs. to grains? " 57600 grs. 3. Bring 8 lbs. to drams^ " 768 drs. 4. Bring 28 drams to scruples? " 84 scruples. 5. Bring 72 scruples to grains? " 1440 grs. 6. Bring 15 lbs. 9 oz. 4 dr. 2 scr. to grs? " 91000 " RE DUCT I 109 TROY WEIGHT. First form of the Table. By this weight, gold, jewels, silver and liquors are weighed. 1 J ounces make 1 lb. 20 pennyweights make 1 oz. 24 grains " 1 dwt. (pennyweight.) Q. How are these numbers used? A. As multipliers, when arranged agreeably to the first form of the table. Second form of the Table. 24- grains make (pennyweight) 1 dwt. 20 pennyweights make 1 oz. 12 ounces (i 1 lb. CASE [. — Bringing a higher name to a lower. APPLICATION. 1. Bring 27 lbs. 10 ozs. 13 dwts. of gold, to grains I Ans. 160632 2. Bring 8 lbs. Ooz. 7 dwts. 2 grs. to grains? " 46250 3. Bring 375 lbs. 10 ozs. 16 " " Ans. 2164816 4. Bring 29 ozs. 16 dwts. to dwts.? " 596 5. Bring 19 lbs. 11 ozs. 14 dwts. 21 grs. to grains? Ans. 115077 6. Bring 125 lbs. to grains? « 720000 7. Bring 59 lbs. 13 dwts. 5 grs. to grains? " 340157 LONG MEASURE. This measure is used for measuring distances. First form of the Table. 360 degrees is equal to the circumference of the earth. 60 J statute or geographic miles, 1 degree. { mi hs (league) - ... i league. 8 furlongs (or 1760 yards,) - - 1 mile, perches (or 220 yards,) - - 1 furlong. 5 J yards, 1 per ch. 10 110 REDUCTfOff. 9 feet, .... 1 yarcfr 11 inches, ... 1 foot. 8 barley corns, ... 1 inch. Second form of the Table, 3 barley corns, - - - 1 inch. 12 inches, - 1 foot. B feet, - 1 yard. 5$ yards, - 1 perch. 40 poles or perches, - 1 furlong. 8 furlongs, - 1 Hi: 3 miles, - 1 league. 60 geographic or 69 J statue miles, 1 deu 360 degrees the circumference of the earth i. Q. How do you bring miles to barley corns, agreeably to RuU 1? A. Miles are multiplied by 8 to reduce them to furlongs, Furlongs u by 40 are perches. Perches " by 5 J are yards, ' rds " by 3 are feet. Feet " by 12 are inches. Inches " by 3 are barlej corns. CASE I. — Bringing a higher name to a lov APPLICATION. 1. Bring 273 miles to inches? Ans. 17297280 in. J. Reduce 2 m. 1 ft. 8 p. 3 yds. 2 in. to inches? Ans. 136334 in. 4. Reduce 29 miles to inches? " 1847440 in. 4. Reduce 16 furlongs to poles? " (J40 ">. Reduce 70 miles, 7 fur. to furlongs? Ans. 567 ft. 6. Reduce 3 leagues to poles? " 2880 ps. 7. Reduce 2 yards, 3 inches to inches? " 75 in. 8. Reduce 17 yards, 1 loot to feet? " 52 ft 9. Reduce 52 perches to yards? " 286 yds. REDUCTION. 1 1 1 CLOTH MEASURE. lenominations of Cloth Measure are English Ell, tish Ell, yard and nail. First form of the Table. 6 quarters make 1 Ell French. 5 " " 1 Ell English. 4 " " 1 yard. 3 « « 1 Ell Flemish. 4 nails " 1 quarter of a yard. Q. How are the numbers used in the form of the above tabic? A. As multipliers. Q. When they are employed as multipliers, what is the reduction called? A. Reduction descending. Second form of the Table. 4 nails (na.) make 1 quarter of a yard. 4 quarters " 1 yard. " " (E. F.) 1 Ell Flemish. 5 " « (E. E.) 1 Ell English. 6 « " (E. Fr.) 1 Ell French. * CASE I. APPLICATION. 1 . Bring 56 English Ells to quarters? Bring 85 French Ells to quarters? Briog 91 Flemish Ells to quarters? \. Bring 25 yards to nails? 5. Bring 27 yards to quarters? 6. Bring 35 quarters to nails? Lns 280 qr. M 510 « u 273 " « 400 na. M 108 qr. ii 140 na. 1 1 2 REDUCTION. LAND OR SQUARE MEASURE. The denominations of Land Measure, are acre, rood, re perch, and square yard. First form of the Table. 4 roods make - . - 1 acre. 40 square perches make - 1 rood. 30i square yards ** _-' - - 1 square foot. ,9 square feet " 1 " yard. 1 14 square inches " - 1 " foot. Second form of the Table. 144 square inches make - - 1 square foot. u feet " - 1 « yard. 301 " yards " - - 1 « perch. 40 " perches " 1 rood. 4 roods " - - 1 acre. CASE I. — Bringing a higher name to a lower, APPLICATION. 1. Bring 35 acres to perches] Ans. 5600p. Bring 30ac. lrd. lOpr. to perches? " 4850p. 3. Bring 5ac. 3rds. 32pr. to perches? " 952p. 4. Bring 1 acre to square yards? Ans. 4840 sq. yd. 5. Bring 1 acre to square feet? " 43560 sq. ft. 6. Bring 15 roods to sq. perches? " # 600 perch. 7. Bring 50 square yards to sq. ft? " 450 sq. ft. 8. Bring 5 square feet to sq. in.? il 720 9. Bring 16 sq. perches tosq. yds? u 484 10. Bring 3rds. 25pr. to sq. inches? " 5684580 sq. in. LIQUID MEASURE. The denominations are gills, pints, quarts, gallons, hogsheads, and tuns. 4 gills make - - - 1 pint. 2 pints Ci - 1 quart. 4 quarts " - - - 1 gallon. REDUCTION. 113 31 \ gallons makt - M a 63 " " 4 hogsheads make 1 barrel. 1 tierce. 1 hogshead. 1 tun. Ans u Note. — 1 pipe is equal to 2 hogsheads; therefore, 2 pipes ,ual to 1 tun. DEDUCTION DESCENDING i CASE I. Tuns multiplied by 4 are hogsheads. Hhds. " by 6*3 are gallons. Gallons " by 4- are quarts. Quarts " by 2 are pints. 1. Bring 17 quarts to pints? Bring 25 gallons to quarts? 3. Bring 5 hhds. to gallons? 4. Bring 100 gallons to pints? In 6 tuns, how many pints? 6. Bring 7 hhds. 41 gals. 2 qrts. to qrts? 7. Bring 47 gals. 2 qrts. to pints? Bring 4 hhds. 3 qrts. to pints? '. Reduce 19 tuns, 27 gals, to quarts? 10. Bring 5 tuns, 1 hhd. 15 gals, to qts? REDUCTION ASCENDING, (iills divided by 4 are pints. Pints " by 2 are quarts. " by 4 are gallons. •Gallons " by 63 are hhds. Hhds. " by 4 are tuns. AVOIRDUPOIS WEIGHT. CASE II. — Bringing a lower name to a higher. (^. How is the operation performed? A. By division. Ans. 34 pt. « 100 qr. " 315 yds. " 800 pt. 12096 pt. 1930 qr: 380 pt. 2022 " 1926.0 qts. 10707 pts. 114 REDUCTION. Q. How are the divisors arranged? A. Agreeably to the second form of the table of Avoir- dupois weight. APPLICATION. 1. Bring 1040 quarters to cwts? Ans. 260 cwt. Bring 1008 pounds to quarters? * 36 qrs. Bring 320 drams to ounces? " 20 ozs. 4. Bring 1130 ounces to a higher denomination? Ans. 2 qrs. 14 lbs. 10 ozs. 5. Bring 540 cwts. to tons? Ans. 27 tons. 6. Bring 7000 lbs. to a higher denomination? Ans. 62 cwt. 2 qrs. 7. Bring 180 quarters to tons? " 2 tons, 5 cwt. Bring 640 drams to ounces? " 40 ozs. 9. Bring 1561 lbs. to cwts. qrs. and Ans. 13 cwt. 3 qrs. 21 lbs. APOTHECARIES WEIGHT. CASE II. — Bringing a lower name to a higher. Q. Recite the second form of the table. A. Grains divided by 20 are scruples. Scruples H by 3 are drams. Drams " by 8 are ounces. Ounces " by 12 are pounds. APPLICATION, 1. Bring 672 ounces to pounds? 2. Bring 1136 drams to ounces? 3. Bring 252 scruples to drams? 4. Bring 320 grains to scruples? 5. Bring 192 scruples to ounces* 6. Bring 80640 grains to pounds? 7. Bring 317 grains to scruples? Ans. 15 scr. 17 grs. 8. Bring 406 ounces to pounds? " 33 lbs. 10 ozs. 9. Bring 4020 grains to scruples? ", 201 scr. 10. Bring 157 drams to ounces? <c 19 ozs. 5 drs„ Ans. 56 lbs. u 142 ozs. u 84 dr. U 16 scr. u 8 ozs. a 14 lbs. REDUCTI 115 troy ^vr.K.in. CASE II. Recite the table referring to this case? \ . Grains divided by 24 are dwts. lYnny weights M by 20 are ounces. Ounces " by 12 are pounds. APPLICATION 1 . Bang 1S72 grains to pennyweights? Ans.78 dwts. •J. Reduec 2320 dwts. to ounces? " 116 ozs. 1000 grains to pounds? " 25 lbs. 4. Reduce 595 pennyweights to a higher denomination? Ans. 29 ozs. 15 dwt.-. . Reduce 175 dwts. to ounces? " 8 ozs. 15 « Keduce 94 ounces to pounds? " 7 lbs. 10 ozs. 7. Reduce 1721 grains to dwts.F « 71 dwts. 17 grs. 8. Reduce 86400 grams to lbs ? " 15 lbs Reduce 5749 dwts. to a higher denomination? Ans. 287 ozs. 9 dwts. 10. In 2385 grains, how many spoons, each weighing 6 Ans. ir> 11. In 160632 grains of pure gold, how many lbs , ozs., and dv . 27 lbs. 10 oz. 13 dwts. LOUG mi: am uk. « \m: ii. te the second form of the table? \ Btrley corns divided by 3 are inches. Inc " by 12 are feet. " by 3 are yards. " by 5 J are pert r poles " by 40 are furlongs. • Furlongs " by 8 are miles. Miles " by 3 are leagues. Also, bj|m " by 60 are geographic degrees- 116 REDUCTION. APPLICATION. 1. In 1837440 inches, how many miles? Ans. 29m. 2. In 19753 yards, how many furlongs? Ans. 89 fur, 173 yds- & In 590057 inches, how many leagues? Ans. 3 lea. 2 fur. 1 10 yds. 1 ft. 5 in. 4. In 22800 barley corns, how many miles? Ans. 11 m. 7ft. 38 p. 2 yds. 2 It. 5. In B4594560 inches, how many miles? Ans. 546m. rches, how many furlongs? M 18 fur. 7. In 144 feet, how many yards? " 48 yds. 8. In 1140 1^0 inches, how many leagues? " 6 lea. :;>200 yards, how many mil " 20 m. CLOTH MEASURE. CASE II. Q. Recite the table corresponding to this case? A. Nails divided by 4 are quarters. Quarters " by 4- are yards. Q. Bring nails to English I A. Nails divided by 4 are quarters. Quarters " by 5 are English Ells. Q. Bring nails to Flemish Ells? A. Nails divided by 4 are quarters. Quarters " by 3 are Flemish Ells. Q. Bring nails to French Ells? A. Nails divided by 4 are quarters. Quarters " by 6 are French Ells. . APPLICATION. 1. Reduce 36 quarters to yards? Ans. 9 yds. 2. Reduce 180 quarters to Flemish Ells? " 60 E. F. 3. Redifce 95 quarters to English Ells? " 19 E. E. 4. Reduce 126 quarters to French Ells? * 21 E. Fr. 5. Reduce 318 nails to a higher denomination? Ans. 19 yds. 3 qr. 2 na. ■6. Reduce 2528 nails to yards? Ans. 158 yds. An* :. 17 qts. u 25 gals. a 5 hhds, u 800 gals. M 6 tuns. REDUCTION. 117 LIQUID MEASURE. CASE II. I ) . Recite the table agreeing with case Second? A. Pints divided by 2 are quarts. Quarts " by 4 are gallons. Gallons " by 63 are hogsheads. Hhds. H by 4 are tuns. APPLICATION. 1 . How many quarts in 34 pints? 2. How many gallons in 100 quarts? .*. How many hhds. in 315 gallons? 4. How many gallons in 800 pints? How many tuns in 12096 pints? 6. How many hogsheads in 1930 pints? Ana. 7 hhds. 41 gals. 2 qts. 7. In 10707 pints, how many tuns, hogsheads, gallons, quart! and pints? Ans. 5 ts. 1 hhd. 15 gals. 1 qt. 1 pt. PROMISCUOUS EXAMPLES. 1 . In 5 hhds. of wine, how many pints? J. hi J7 hhds. 16 gallons, how many quarts* 3. In 942 tuns, how many pints? 4. In 50400 pints, how many tuns? 5. In 3240 half pints, how many hogsheads, gallons, and quarts? TIME. The denominations are seconds, minutes, hours, days, wi t ks, months, and years. CASE I. — To bring years to seconds. tt multiplied by 12 are months. Months " by 4 are weeks. Weeks " by 7 are days. DayB « by 24 are hours. Hours " by 60 are minutes. Minutes " by 60 are seconds. 118 REDUCTION. APPLICATION. 1. Bring 1 year, 10 months, 2 weeks, 3 days, 12 hours 10 min. 15 sec. to seconds? Ans. 314-07015 sec. J. Bring 30 minutes to seconds? '« 1800 3. Bring 12 years to months? " 144 4. Bring 3 days, 5 hours, 29 minutes to minutes? Ans. 4649 min. 5. Bring 37 weeks, 5 days to minutes? * 380160 CASE II. Q. Recite the table corresponding to this case? A. Seconds divided by 60 are minutes, utes " by 60 are hours. Hours " by 24 are days Days " by 7 are weeks. Weeks " by 4 are months. Months " by 12 aie years. LAND'OR SQUARE MEASURE. CASK II. APPLICATION. 1. Reduce 160 roods to acres? Ans. 40 acres. 2. Reduce 487 sq. yds. to sq. perches? " 28 sq.pr. 3. Reduce 878 sq. ft. to square yards? " 42 sq. yds. 4. Reduce 600 perches to aci- " 3 a. 3 rds. 5. In 5600 perchesy»how many acres? il 35 acres. & In 4850 perches, how many acres? " 30a. lr.lOp 7. In 484 sq. yds. how many sq. pr.? M 16 8. In 5684580 sq. inches, how many roods and perches? Ans. 3rds. 25 perches. In 576 sq. inches, how many sq. feet? Ans. 4 10. In 115520 sq. inches, how many sq. yds.? U 120 11. In 37974 square inches, how many square yards, square .feet, and square inches. Ans. 29 sq. yds. 2 sq. ft. 102 sq. in. 12, In 3946 perches, how many acres, roods, and perches? Ans, 24 a. 3 rds. 26 ps. REDUCTION. 119 DRY MEASURE. The denominations are 2 pints make 1 quart, 8 quarts one peck, and 4 pecks one bushel. Reduction Descendini:. Q. How do you bring bushels to pints? Rile 1. — Bushels multiplied by 4 are pecks. Pecks " by 8 are quarts. Quarts <£ by 2 are pints. 1. In 1 bushels, how many pints? I 16 pecks. 8 128 quarts. 2 ••J pints. Rule 2. — Pints divided by 2 are quarts; quarts divided by 8 are pecks; pecks divided by 4 are bushels. APPLICATION. 1 . Bring 32 pecks to quarts? Ans. 256 qts. 2. Reduce 7 bushels to pecks? «• 28 p. & Bring 12 bushels to pints? " 768 pts. . 4. Bring 15 bushels, 3 pecks, to pecks? " 63 p. Bring 25 bushels, 1 peck, 2 quarts, 1 pint, to pints? Ans. 1621 pts. MOTION OR CIRCLE MEASURE. The denominations are — 60 seconds make 1 minute, v 60 minutes " 1 degree, 30 degrees " 1 sign, sig. 12 signs " 1 revolution or circle. 120 REDUCTION. The 12 signs of the Zodiac are as follows: •dries. Taurus. Gemini. Cancer. Leo. Virgo. Li lira. Srorpin. Saggitarius. Capricomus. Aquarius. Pisces. APPLICATION. Reduce 24° to minutes? In 5 signs, how many seconds? Reduce 11 sig. 12° to degrees? Ans. 1440 Ans. 540000 4 < 342° 4. How many second are in 4 sig. 3° 18m. 27 seconds? Ans. 443907^ NAMES OF THE TWELVE CALENDAR MONTHS. January, 1st month has 31 days. February, 2d month has 28, (except leap year, which gives it 29.) March, 3d month has 31 c layi April, 4th u 30 u May, 5th a 31 it June, 6th tc 30 a July, 7th a 31 u August, 8th u 31 a September, 9th « 30 a October, 10th tf 31 a November, 11th M 30 u December, 1-Jth it 31 u OTHERWISE EXPRESSED. The fourth, tlevcnth, ninth, and sixth, Have 30 days to each affixed, And every other 31, Except the second month (February) alone, Which has bu{ twenty-eight in fine, Till leap year gives it twenty-nine. Note. — Every fourth year is leap year. Q. How many days in a common year? A. Three hundred and sixty-five days, and six hours. REDUCTION. 121 Q. How many days in a leap year? A. Three hundred and sixty-six. ILLUSTRATION. 365 days, 6 hours, in a common year. 366 " in leap year. So that 4 X G = 24 hours, or 1 day, gained every fourth year, which is called leap year. QUESTIONS FOR EXERCISE IN REDUCTION. 1. In 40 dollars, how many cent.>! Ans. 4000. In 96 furlongs, how many miles? " \2. 3. In 245 days, how many weeks? " 35. I. In <i tons, 15 cwt. how many cwts.? u 135. 5. In 96 half cents, how many cents? " 48. ti. In '27r20 perches, how many roods? " 63. 7. In 176 pecks, how many bushels] *< 44. 8. In 108 pints, how many quarts? « 54. 9. In 720 seconds, how many minutes? " 12. 10. In 474 gals, how many hhds? Ans. 7 hhd. 33 gals. 11. In 103 pints, how maDy quarts? " 51 qts. 1 pt. 12. In 273 miles, how many leagues? Ans. 91. 1 '{. In 108 dwts. how many ounces? Ans. 5 oz. 8 dwts. ! 1. In 250 shillings, how many Ls? " 12 Ls. 10 s. 15. In 21 scruples, how many drams? Ans. 63. !»>'. 1 n 203 days, how many weeks? " 29. 17. In 64 nails, how many yards? " ].. 18. In 84 drams, Avoirdupois weight, how many ounces? Ans. 5 ozs. 4 drs. 1!>. In 4320 minutes, how many days? Ans. 3. "> cut. how many tons? " 20. In :i2 furlongs, how many mi: « 4. 22. In l'J%0grs. ofgold,how many lbs? Ans.21bs. 3oz. 23. In 13 lbs. Avoirdupois, how many drs? Ans. 3328. •J I. In 1 ton, how many pounds? " 2&40. What will 1 ton of iron come to at 61 cts. per lb.? 11 Ans. $140.00. 1 29 REDUCTION. 26. In 27 lbs. 6 ozs. gold, how many grs? Ans. 1 584-00/ 27. In 252 bushels, how many quarts? " 8064. 28. In 8064- quarts, how many bushels? " 252. 29. In 484-0 sq. yds. how many sq. per.? " 160. 30. In 156 qrs. how many Ells Flemish? " 52. 31. In 760 " how many Ells English? « 152. . In 840 " how many Ells French? « 168. 33. In 27552 lbs. how many cwts.? " 984. 34. In 584621 gallons, how many tons? Ans. 9279 tons, 3 hhds. 44 gals. 35. In 86400 seconds, how many days? An>. I. 36. Bring 1 mile to barley corns? Ans. 190080. Bring 1 ton to drams? 38. Bring 5 lbs. Troy weight to grains? 39. Bring 5 lbs. Avoirdupois wt. to drs.r 40. Bring 5 lbs. Apothecaries wt to 41. Bring 5 signs of the Zodiac to sec? . Bring 3 ds. 5 hrs. 29 m. to minutes? 43. Bring 37 yards, 1 foot, to feet? 44. Bring 880 pints to gallons? 45. In 1 hhd. of ale, how many pints? 46. In 2310 pints, how many bushels? Ans. 36 bushs. p. 3 qts. 47. In 4032 qts. of bush, how many hhds? Ans. 16. 48. In 36 bu. 3 qts. of corn, how many pts.? " 2310. 49. In 6888 hours, how many weeks? Ans. 41. 50. In 41 weeks, how many hour^r " 6888. 51. In 2850 perches, how many acres? Ans. 17 acres, 3 roods, 10 perches. 52. In 17a. 3r. lOp. how many perches? Ans. 2850. 53. In 72 leagues, how many yards? Ans. 380160. I 54. In 3 tons of lead, how many lbs.? " 6720. 55. In 7 miles, how many feet? " 36960. 56. In 484 yards, how many perches? " 88. 57. In 2400 English sovereigns, how many dollars? (allowing the value of a 'sovereign' to be $4.80.) Ans. $11520. it 57344a If 28800. « 1280. H 28800. u 540000. tt 4649. 14 112. K 110. II 504. 123 58. In 70 boxes of sugar, each 12 lbs. how many cwts.? Ans. 7 cwt. 2 qrs. 10 sq. in. how many acres? Ans. 1. In 1000 dimes, how many Eagles? " 10. In 423 barley corns, how many inches? " 141. 62. In 820 cwts., how many tons? « 41. In 100 dollars, how many half dimes? Ans. 2000. In 5280 feet, how many miles? " 1. 65. How many times does a regular clock strike in a days? Ans. 156. . How many seconds in a year, allowing it to be 365 days, 6 hours? Ans. 31557600. CUBIC OR SOLID MEASURE. REPEAT THE TABLE. 1728 solid or cubic inches make 1 solid foot. 40 feet of round timber, or ) , - . 50 feet of hewn timber, ) ALSO, 27 solid or cubic feet make 1 yard. 1 28 solid feet, or 8 feet long, ) , , , c . * feet wide, and 4 ft. highlf make ' «">rfofwood. 1. If a pile of wood be 40 feet long, 4 feet wide, and 4 feet high, how many cords are therein? Ans. 5 cords. J. Suppose a pile of wood is 48 feet long, 5 feet wide, and 3 \ feet high, how many cords in it? Ans. 6 T * ff cds. Q. How is the operation performed? A. By multiplying the length, width, and height to- • r, and dividing the product by 128 for the number The cellar of a house is 20 feet long, 30 wide, heighth or depth 10 feet, what will the digging thereof come to at per cubic yard? Ans. $15.55$. Q. How is the above question done? A. By multiplying the length, breadth, and depth together, and dividing by 27 for the number of cubic yards we get 222f cubic yds. at 7c. per is $15.55f. 124 REDUCTION. Q. What is a cubic foot? A. It is 12 inches long, 12 inches broad, and 12 inches deep. 12 x 12 -f 12 = 1728 solid inches. Q What is a cubic yard? A. In length it is 3 feet, in breadth 3 feet, and in depth 3 feet; consequently, 3 x. 3 X 3 = 27 feet in a solid yard, 3 3 = 27. Q. How many feet in a square yard? A. Nine feet, 3x3 = 9 feet in a square yard. Q. How is a number squared? A. By multiplying the number by itself, as 4* =16, or 4 X 4- = 16. Q. How is the number cubed? A. By multiplying the square of a number by the same number, as 4 X 4 = 16x4 = 64, or 4x4x4 = 16. What is the square oi What is the cube of 5? What is the square of 8? What is the cube of 8? What is the square of 9? What is the cube of 9? 1. How many solid inches in 2 solid feet 5 Ans. 3456. 2. How many solid ft. in 345600 solid in.? " 200 ft. 3. How many solid it. in 691200 solid in.? " 400 ft. 4. In 2 cords of wood, how many solid ft ? " 25(> ft 5. In 30 cords, how many i Ans. 3840 ft. (). A floor in a certain building is 22 {eet long, and 18 feet wide, how many feet of boards will cover the floor 5 Ans. 22 X 18 = 396 ft. 7. A board is 20 feet, 9 inches long, and 14 inches wide, how many feet are contained therein? Ans. 24 ft. 2i in. A >VNOPSIS OF THE PRECEDING RULES OF ARITHMETIC. C3~ The learner should be required to recite the fore- going tables and rules in Reduction, ascending and de- scending. REDUCTION. 125 FEDERAL MONEY. Add 6 07 8 " 3 09 7 " 7 06 3 " 9 03 2 25 27 Q. Why do you prefix a cipher before 7 cents in the first line. A. I must always do so, when the cents are less than 10: thus, 7, 9, 6, and 3, in the column of cents are less than 10, consequently, ciphers must be prefixed thus: 01 for 1 cent, 02 for 2 cents, 03 for 3 cents, 04 for 4- cents, 05 for 5 cents, 06 for 6 cents, 07 for 7 cents, and soon up to 10. If I buy 4 lbs. of coffee for 50 cents, 3 lbs. of tea for $2.50, 7 lbs. candles for 87$ cents, and 1 gallon of wine, for $1,935 cents, what must I pay for them? Coffee, - $0.50 Tea, - - - 2.50 Candles, - - - 0,871 Wine, - ' 1935 My cook has bought in market a turkey for $1.87$ cts. a pair of ducks for $1.18|, a quarter of lamb for 432 cents, a quarter of veal for $1.37$, a piece of beef for 68| lj and a peck of apples for 12$ cents, what sum must I irive to pay for the articles] , A turkey, - - - $1.87$ Pair of ducks, - - 1.18| Quarter of lamb, - - 432 Quarter of veal, - - 1.37$ of, 68| Apples, - - - 12$ 11* 126 REDUCTION. ADDITION AND SUBTRACTION OF FEDERAL MONEY. 1. If from $100.00, there be paid atone time $17.28$, at another time $1000$, and another time $37.15, how much will remain? Ans. $35.56. 2. Subtract £ cent from $100. 3. Subtract f of a cent from $100. 4. Subtract 87* cents from $50.06*. 5. Subtract 7 cents from $20. 6. Subtract 1 £ cents from $20. 7. Subtract 9| cents from $104,061. 8. Take * of a cent from $100. ENGLISH OR STERLING MONEY. The "denominations are farthings, pence, shillings, and pounds. 4 farthings make - - - 1 penny, d. 12 pence " 1 shilling, s. 20 shillings " - 1 pound, L. This standard is said to have been fixed in the reign of Richard 1st, by persons from the Eastern parts of Ger- many, called EasterlingSj and hence, the word sterling, which is now applied to all lawful money of Great Britain. Li «• d. L. t. d. Example 1. 2 3 4. Example 2. 7 2 4* 7 12 13 7 6J 9 7 3 4 5 2 5 2 2j 10 18 10| .€23 13 11$, £35 14 Rule. — Add up the farthings, divide the sum by 4, and if there be a remainder, set down the remainder as so many farthings, carry the quotient to the pence column, and add them up as in common addition, and divide the sum by 12, because 12 pence is a shilling, if there be a remainder set it down under the pence column, and car- ry the quotient to the column of shillings. In like man- ner add the shillings, and divide the sum by 20, for REDUCTION. ! 21 pounds, because 20s. is a pound; if there be a remainder set it down under the column for shillings, and carry the quotient to the pounds, the pounds added up above 10 as in common addition, will give the result required. REDUCTION DESCENDING. Rule 1. — Pounds multiplied by 20 are shillings, shil- multiplied by 12 are pence, pence multiplied by 4, are farthings. L. $. d. Bring 10 17 6f to farthings. 20 217 12 2610 4 10443 farthings. L. $. d. Bring 68 16 6$ to farthings. Bring 17 15 3} to farthings. Bring 25 10 S\ to farthings. Bring 100 19 9| to farthings. Bring 19 19 11| to farthings. (^. What are farthings added above? Ans. 4. U hat are pence added above? " 12. (^. What are shillings added above? " 20. (^. What are pounds added above? " 10. Q. How do you bring pounds, shillings, pence and farthings to farthings? A. The pounds I multiply by 20, by Rule 1st, and add in the shillings, (if any,) the shillings I multiply by 12, to bring them to pence, the pence I multiply by 4, to bring them to farthings. REDUCTION ASCENDING. Q. How is this kind of Reduction performed? A. By Division. REDUCTION. Q. What is the rule, Rule 2d. A. Farthings divided by 4 are pence, d. Pence " by 12 are shillings, s. Shillings, " by 20 are pounds, L. Example. — Bring 30096 farthings to pounds. Bring 96000 farthings to pounds. Bring 19199 farthings to pounds. PROMISCUOUS EXAMPLES IN STERLING MONEY. L. p. d. L. ». d. Add 183 19 81 Add 648 17 6 a 39678 15 4i " 365 14 9 u 84374 15 3| " 487 17 8i it 94287 17 4$ « 935 15 91 a 93687 16 9f « 787 17 6f SUBTRACTION OF STERLING MONEY. From 1008 19 6f Take 784 18 71 Rem'r. 224 00 11J < / '4 1 1 1 j B i) Q. How is this rule performed? Rule.— As 4 farthings make 1 penny, 4 will be a com- mon denominator, and in the above example, where 7 pence is the subtrahend, and 6 pence the minuend, you take 7 from 12, and the remainder is 5, which added to 6 in the minuend make 11 pence, which set down as the REDUCTION. difference in the pence column, then carry 1, because speak of 12 to the shillings, 1 carried to 18, make 1 9 from 1 9 leaves 0: then subtract the pounds, as in inon subtraction. ,m 1998 17 0| From 198 17 8\ Take 1348 15 9i Take 139 19 llf I have a purse of money containing .£100 2s. 4>±s, if I take out £60 7*. Sid. what sum will be left? Ans. £939 14*. 7|</. FEDERAL MONEY. Lent a man $400, he now returns $211.12$ cts. how much is unpaid? Ans. $188.87$. My carpenter's bill rendered this day is $110.95| cts. I paid him $90.10|, how much is the balance due him? Ans $20.84$. ADDITION OF COMPOUND NUMBERS. Q. What is the object of adding compound numbers? A. To unite parts of the same denomination in such a manner that their units may stand under each other. Q. Where do you begin to add? A. At the right hand column. Q. Why so? A. Because the addition of compound numbers de- pends on the same principles, as, that of simple numbers. <<!• How do you proceed? A. By adding separately the sum of each column, al- s recollecting how many parts of each denomination it takes to make one of the next higher. Q. By what do you divide the amount? A. By as many of that denomination as will make one of the next greater, as before. Q. Ann dividing) ii there should be a remainder, how do you proceed? A. The remainder is set down, and the quotient pro- 130 REDUCTION. duced by dividing, carried to the next highest denomina- tion. In this manner, the process is continued until all the denominations have been added up. TROY WEIGHT. lbs. or*. dwts. grs. lbs. ozs. diets, grs. Add 17 3 15 11 Add 17 3 15 17 u 13 2 13 13 " 11- 2 17 13 M 15 3 14 14 " 19 7 15 19 a 13 10 " 18 6 17 13 U 12 1 17 " 18 5 18 22 (( 13 14 71 9 17 21 AVOIRDUPOIS WEIGHT. ewts. grt. Of. ozs. drs. tons. acts, qrt . lbs. ozs. dr. Add 15 2 15 15 15 Add i 13 8 7 u o 3 3 14 13 a 1 16 10 6 « 12 2 13 14 14 M 5 14 2 17 6 8 « 10 1 17 15 a 9 11 3 18 5 13 M 12 1 10 10 a 14 12 1 11 10 14 " 13 2 17 18 IS (w 1 23 10 APOTHECARIES WEIGHT. lbs. ozs. drs. scr. grt. lbs. ozs. drs. scr. grs. Add 12 10 5 2 18 Add 8 9 4 1 If) • s 4 7 111 " 10 11 7 2 14 " 9 8 7 2 10 " 14 8 5 2 15 LONG MEASURE lea. ms.fur. pr. yds. A in. dd 5 2 4 17 Add 16 2 11 u 16 1 3 10 a 1 1 9 a 7 -J n .") 24 a 14 2 11 u 526 3 12 a 99 1 8 a 834 2 6 34 a 108 2* 10 a 38 3 12 <( 436 2 7 1493 OK. 131 CLOTH URE. yd*. qrs. n. E. E. qrs. n. Add 75 3 2 Add 2 .. 163 l 8 a 98 2 3 u 2 1 5 •J a 47 1 2 a 738 3 1 a 96 2 2 « 1786 2 3 « 149 1 3 3009 1 1 LAND MEASURE. 4 » acs. ni». pr. acs. rds. pr. Add 39 2 37 Add 4968 3 27 « 1 17 a 9484 2 32 M 68 38 « 9694 1 26 U 129 3 12 c< 4947 3 34 CI 532 ! Is 832 2 2 LIQUID MEASURE, ton*, hhds. gah. hhds. gals. qts. pts. Add 18 2 Add 346 42 3 1 " b3 1 39 " 327 04 " 46 1 19 ' « 285 3 28 a 27 36 1 o U 468 24 1 o it 543 37 2 1 u 964 36 3 1 741 1 18 DRY MEASURE. totsA. />A-s. y/j. bush. pks. qts. pts. Add 37 2 1 Add 42 1 5 1 u 182 3 2 a 63 2 3 M 422 1 (C 44 3 7 u 162 3 1 a 65 2 6 1 <t 7 2 I G 132 REDUCTION. TIME. yrt. ms. tos. ds. hrs. hrs. ms. tec. Add 17 11 3 5 20 Add 20 52 40 " 172 9 2 3 17 " 122 12 35 " 35 7 3 6 22 " 68 9 17 « 4 10 4 16 « 135 17 12 " 6 3 19 « 24 35 28 231 6 3 3 22 MOTION OR CIRCLE MEASURE. • sign, dgs. mins. sect. sign, degs . m. sees. Add 1 5 37 42 Add 1 4 37 42 M 1 7 26 12 " 5 4 44 48 N 4 8 26 .11 9 12 48 40 M 1 4 82 17 9 14 42 48 M 3 6 47 v 11 2 3 9 MULTIPLICATION OF COMPOUND NUMBERS. Q. What is the utility of compound multiplication? A. To find the product of numbers, one of which is compound, and the other, a simple factor. RULE. Multiply the compound quantity by the simple multi- plier, beginning with the lowest denomination of the multiplicand; this done, divide the product by the number which it takes to make one of the next superior, putting down the remainder, (if any,) we then add the quotient last produced to the product of the next denomination by the multiplier to reduce this sum, putting down the re- mainder as before, and proceed in this manner through all the denominations to the last, which is to be multi- plied like a simple number. ILLUSTRATION OF THE RULE. Suppose we multiply 4 cwt. 2 qrs. 16 lbs. by 9, we be- gin with the lowest denomination, and multiply 16 lbs. by REDUCTION. 133 9, and find the product is 1 1 !• lbs. ore divide 1 Vi lbs. by 28, L» ^ 28 lbs. to make 1 quarter, which is one of the next superior, and lind the quotient is 5, and a remainder of 4, which we set down. We next multi- ply the nexl highest number 2 quarters by 9 'adding in' i quotient, and we get 23 quarters; we reduce this sum by dividing it by 4, because 4 quarters make 1 cwt. and set down the remainder 3, which is 3 quarters, and proceed in the same manner, always obsorving that the last number is to be multiplied like a simple number. Note. — When the multiplier exceeds 12, that is, when it ii so large that it may be inconvenient to multiply by the whole at once. The shortest method is to resolve it into factors, if it can be done — if not, into powers of 10, or thai: FEDERAL MONEY. Multiply $104 33 3 9 $938 99 7 same as decimals. or thus: Multiply $104 33 3 10 1048 33 104 33 3 By adding a cipher minus the given number as be- $938 99 7 fore. ENGLISH MONEY. L. s. d. L. s. d. Multiply 12 3 65 by 8? Multiply 9 11 4* by 5? AVOIRDUPOIS WEIGHT. qr». lbs. cwt. qrs. lbs. Multiply 1 2 6 by 70? Multiply 1 3 14 by 21? TROY WEIGHT. /6s. ozs. dirts, gr. lbs. ozs. dxrts. grs. Multiply 67 5 16 1 by 2? Multiply 17 9 14 2 by 10? IS 1 3 i REDUCTION- APOTHECARIES WEIGHT. lbs. ozs. drs. scr. grs. Multiply - 12 10 5 2 18 by 5? " 16 6 6 2 18 by 4? LONG MEASURE. lea. m. fur. pr. yds. ft. in. b. c. Multiply 5 2 4 17 by 5] Mul. 16 2 11 2 by 6? CLOTH MEASURE. yds. qrs. ns. yds. qrs. ns. Multiply 75 3 2 by 12? Multiply 35 2 2 by 15? TIME. yrs. m. to. d. h. Multiply 17 11 3 5 20 by 8? " 6 3 19 by 5? LAND MEASURE. act. rds. pr. acs. rds. pr. Multiply 39 2 25 by 8? Multiply 45- 3 32 by 5? LIQUID MEASURE. tuns. hhds. gals. turn. hhih. Multiply 18 2 54 by 10? Multiply 6 1 23 by 8? DRY MEASURE. bush. pks. qts. bush. pks. (jts. Multiply 37 2 1 by 8? Multiply 44 1 5 by 10? MOTION OR CIRCLE MEASURE. sign, deg. ms. sec. sign, deg. ms. sec. Multiply 1 5 37 42 by 5? Mul. 3 8 22 25 by 10? SUBTRACTION OF COMPOUND NUMBERS. This operation is performed in the same way as the subtraction of whole numbers, except with regard to the number, which it is necessary 'to borrow' from the higher denominations, in order to perform the partial subtrac- tions, when the lower number exceeds the upper. AVOIRDUPOIS WEIGHT. cwts. qrs. ozs. lbs. drs. cwts. qrs. ozs. lbs. drs. From 22 1 7 6 13 From 5 17 5 9 Take 13 8 8 14 Take 8 3 21 1 7 9 26 13 15 REDUCTION. 135 TROY WEIGHT. lbs. ozs. dirts, grs. lbs. ozs. dwts. grs. From 17 9 H 2 From 7 3 14, 11 Take H 5 14 18 Take 3 7 15 20 APOTHECARIES WEIGHT. lbs. ozs. drs. scr. grt. lbs. ozs. drs. scr. grs. From 12 10 5 2 18 From 8 4, 6 2 16 Take 6 9 6 2 18 Take 4, 8 7 1 19 LONG MEASURE. Ua. ms.Jxtr. pr. m. fur. pr. yds. From 9 2 7 17 From 10 5 20 4 Take 6 1 6 25 Take 8 7 35 5 CLOTH MEASURE. yds. grs. n. E. E. qrs. n. From 75 3 2 From 75 4 3 Take 35 2 3 Take 63 4 3 TIME. yrs. nu. tcs. dt. hrs. hrs. ms. sec From 17 6 3 6 20 From 20 52 62 Take 12 8 3 5 18 Take 12 58 58 no o i 2 LAND MEASURE. acs. rds. pr. acs. rds. pr. From 45 2 33 From 24 05 Take 39 3 35 Take *J-2 3 35 5 2 38 LIQUID MEASURE. tuns. hhds. gals. qts. pt. hhds. gals. qts. pt. From 18 2 52 2 1 From 6 42 3 1 fain 12 3 58 3 Take 3 33 3 1 5 3 56 3 1 136 'V DRY MEASURE. bush. pks. qts. bush. pks. qts. pt. From U 1 5 From 3 3 3 1 Take 37 2 6 Take 2 3 7 1 6 2 7 MOTION OR CIRCLE MEASURE. sigriy dgs. mins. sees. degs. ms. sees. From 3 8 22 35 From 37 33 35 Take 1 5 37 42 Take 22 35 55 DIVISION OF COMPOUND NUMBERS. A compound number may be divided by a simple num- ber, by regard'ng each of the terms of the foruM forming a distinct dividend. RILE. Divide the highest term of the compound number by the given divisor, reduce the remainder, (if any) to the next lower denomination, adding it to the number of this denomination, and divide the sum by the divisor, reduc- ing the remainder as before, and proceed, in this way through all the denominations to the last. TROY WEIGHT Illustration, 67 5 16 22 by 2. Dividing by 2 is taking i)33 8 18 11 Here 33 is the quotient, and 1 of a remainder, which is 1 lb. this reduced to the next lower denomination which is ounces, we get 12 -f 5 = 17 ounces, this divided by 2, gives 8 in the quotient, and 1 of a remainder, which is 1 oz., this reduced to the next lower, which is penny- weights, and we get 20 -f 16 = 36 pennyweights; 36 divid- ed by 2 gives 18, and no remainder; the next lower de- nomination is 22 grains, and this also divided by 2 gives 11. All other compound quantities are divided upon the same principle. -REDUCTION. 137 LAND MEASURE. act. rds. pr. acs. rds. pr. Divide 45 3 32 by 8? Divide 24 3 05 by 5? CLOTH MEASURE. yds. qrs. n$. yds. qrs. ns. Divide 196 2 2 by 4? Divide 48 3 3 by 61 AVOIRDUPOIS WEIGHT. tons, cwt. qrs. lbs. tons, act. qrs. lbs. Divide 6 14 2 18 by 4? Divide 8 12 3 24 by 8? LIQUID MEASTRE. tuns, hhds. gals. qts. tuns, hhds. gals. qts. Divide 6 4 44 3 by 3? Divide 36 62 2 by 4? DRY MEASURE- bush. pks. qts. bush. pks. qts. Divide 64 2 6 by 4? Divide 84 3 7 by 6? MOTION OR CIRCLE MEASURE. The moon revolves through 12 signs of the Zodiac in J7 days, 7 hours, 42 minutes, 48 seconds. In what time does it describe one sign? Ans. 2d. 6A. 38m, 34s. 12' LECTURE VIII. FRACTIONS SIMPLIFIED. • DEFINITIONS. Q. What is a fraction? A. A fraction is an expression of a part, or some parts of any thing considered as a whole. Q. How is it denoted? A. By two numbers, one placed below the'other, with a line between them as |, written three- fourths. Q. What is the number above the line called? A. It is called a numerator, (as before mentioned) from the French, numei'ateur, which determines the number of parts, it also represents a remainder after division. Q. What is the number below the line called? A. It is called a denominator, from the Latin, denomino, because it denominates the number of parts. Q. What are the numerator and denominator consi- dered? A. They are generally considered the terms of a fraction? Q. How are fractions arranged? A. Into four classes, viz: proper, improper, compound, and mixed. Q. What is a proper fraction? A. It is that, whose numerator is less than the denomi- nator, as I or T ^, &c. Q. What is a complex fraction? A. It is that, which has a fraction in its numerator or denominator, or in both of them: thus, 5$, 8, 4|. 6 94 7. FRACTIONS. 139 Q. What is an improper fraction? A. It is that whose numerator is greater than the de- nominator, as V or J, &c. Q. Whit is a compound fraction? A. It is the fraction of a fraction, or several fractions connected by the proposition of between them, as the J and \ of f of a dollar. Q. What is meant by a mixed number? A. A whole number, and a fraction together, as 3J, or 12|, &c. Q. Can a whole, or integer number be expressed like a fraction? A. Yes, by writing 1 below it as a denominator, as 3 is f, 4 is f or 25 is y. This is evident, because a unit il neither a multiplier or a divisor, that is, you neither in- crease or decrease the value of any thing by multiplying or dividing by 1, as 6 X 1 = 6, f = 6. Q. How is the value of an improper fraction obtained? A. By dividing the numerator by the denominator, as 3, V - *. Q. What is a reciprocal fraction? A. It is a fraction inverted, as | is the reciprocal of \. RULES. I . When the numerator is less than the denominator, the fraction is less than 1, as J or J, &c. 9. When the numerator is equal to it, the fraction is equal to 1, as f or f, &c. .*. When the numerator is greater than the denomina- tor, the fraction is greater than 1, as f or f , &c. Q. Can you tell by inspection when a proper fraction may be less than another, as f , f , \ or J? A. Certainly, | is greater than \ , because the numera- tor does not shew a division of so many parts, for accord- ing to the meaning attached to the words numerator and denominator, it is plain, that a fraction is increased by HO FRACTIONS. increasing its numerator, without changing its denomina- tor, and that a fraction is diminished by diminishing its numerator, without changing its denominator; also, that a fraction is diminished when its denominator is increased without changing its numerator. Q. What is meant by the greatest common measure of two numbers? A. The largest number that will divide them without a remainder, to the lowest terms. Q. When is a fraction in its lowest terms? A. When no number but a unit, will measure both its terms. Q. What is a prime number? A. A number which can only be divided by itself, or a unit, as 7, 9, 13, 19, 5, 23, &c, (as before mentioned in the work.) Q. What is a composite number? A. A number which is equal to the products of its factors, as 28 - 7 x 4, 96 - 6 X 8 X 2 — 96. Q. What is an abstract fraction? A. An abstract fraction is a fraction derived from ano- ther, by means of Reduction. PREPARATORY QUESTK CASE I. I. To find the greatest common measure of two given numbers. RULE. "The greater by the less divide, The less, by what remains beside; The last divisor still again, By what remains, till nought remains; And what divides and leaveth nought, Will be the common measure sought." 1. What is the common measure of 112 and 120? Ans. 8. FRACTION'S. 141 R hat if the greatest commou measure of 379 and Ans. 31. it is the least common measure of 56 and 108? Ans. 4. 4. What is the greatest common measure of 57 and Ans. 57. What is the greatest common measure of 169 and Ans. 1. t is the greatest common measure of 175 and Ans. 35. CASE II. II. To find the least common multiple of any given number. Rule. — If the given numbers are prime to each other, their continued product is their least common multiple. "If not all prime to all beside, See what will two or more divide: Divide the two or more thereby, The same upon the quotients try; And if thou canst divide them do; 'Till nought will measure any two, Then the last quotients multiplied, And all the numbers which divide Continually; the product got, Wilt be the multiple that's sought." 1 . What is the least common multiple of 3, 5, 8 and 10? Operation, 5)3 5 8 10 2)3 1 8 2 3 1 4 3 1 last quotients. 1-2 o divisor. common multiple, 24 5 120 answer. 1 W FRACTIONS. 2. Find the least common multiple between 12, 25, ■ind 42? Ans. 2100. 3. Find the least common multiple of 12, 16, 20, and 30? Ans. 240. 4. What is the least common multiple of 25, 35, 60 and 72? Ans. 12600. 5. What is the least common multiple that will mea- sure 3, 4, 8, and 12? Ans. 24. 6. What number is the least, that 7, 8, 16, and 28, will measure? Ans. 112. CASE IV. To reduce fractional parts of a dollar to cents, Ri i.e. — Multiply the numerator by 100 (because 100 cents is a dollar) and divide by the denominator. 1. Bring | of a dollar to cents? 100 5 8)500 62| or 62 \ cts. f may be reduced to \ because 4 will divide the numerator and denominator without a remainder, thus: 4)| = $ reduced to its lowest terms. How many cents in \ of a dollar? Ans. $.50. 2. Bring £ of a dollar to cents? Ans. $.87$. 3. Bring f of a dollar to c Ans. $.37$, 4. Bring i of a dollar to cen; Ans. $.12$. 5. Bring f or I " to cents? Ans. $.75. 6. What number of cents in j% of a dollar? Ans. $.90. 7. What number of cents in f of a dollar? Ans. $.60. .8. What number of cents in f of a dollar? Ans. $.80. 3E V. To reduce fractions to their lowest denominations, and a/so into cents. 1. Reduce $££ to its lowest terms? Ans. $^ T or 19 Y J T cts. 2. Reduce $§$■ to its lowest terms? Ans. $| or 75 cts. FRACTIONS. 149 3. Reduce $|$ to its lowest t< Ans. $f or 85$ cts. 4. Reduce $if } to its lowest terms? Ans. $}f or 56/, ,'rt to it.s lowest terms? Ans. $f or 40 cts. e $f \ J to its lowest terms? Ans. $ 1 % or 90 cts. 7. Reduce $±H to its lowest terms? Ans. $^ e ot56\ cts. 8. Reduce $$} to its lowest terms? Ans. $f or 60 cts. 9 . Reduce $HJ to its lowest terms? Ans. $£ or 80 cts. K A 1 M OF FRACTIONAL PARTS OF A DOLLAR. 1. What is the ratio between \ and J? Ans. 2. From the preliminary examples it is evident that 2 quarters are = to a half; therefore, the ratio is as 1 to 2 as required. J. Wli.a is the ratio between \ and J? Ans. 3. 3, What is the ratio between \ and f-? Ans. 3. 1. W il ratio between £ and J? Ans. 3. 5. What is the ratio between £ and £? Ans. 7. What is the ratio between T s ff and jf? Ans. 5. USEFUL THEOREMS IN FRACTIONS. Note to Teachers. — The learner should be required to recite these theorems and to apply them practically. THEOREM I. To add or subtract fractions which have the same common denominator, the sum or difference of their numerators must be taken, and the common denominator written under the result. THEOREM II. To reduce fractions to the same denominator, the two I of each of them must be multiplied, by the de- nominator of the other. THEOREM III. A fraction can be multiplied in two ways; namely, by multiplying ill numerator or dividing its denominator. . multiply 3 ' ff by 5 == 3 * which reduced is £ or 3 ' -7- 5 (by dividing the denominator 30 by 5) = «. 1 14 FRACTIONS. THEOREM IV. A fraction can be divided in two ways, by dividing its numerator or multiplying its denominator: thus, divide J by 4 by dividing the numerator we get | and by mul- tiplying the denominator | X 4 =» ^ = | which is exactly the same. THEOREM V. Multiplication alone, according as it is performed on the numerator or denominator, is sufficient for the multi- plication and division of fractions; that is, when you multiply the numeiator you increase, and when you multiply the denominator you decrease, case I. By multip. numerator the fraction is ) . . ,. , By dividing the denominator the fraction is ) P • CASK By dividing the numerator the fraction is ) ,. . , , By multiplying the denominator the fraction is ) THEOREM VI To multiply a whole number by a fraction. Rule. — Multiply the number by the numerator and divide by the denominator: or divide the number by the denominator and multiply the quotient by the numerator. Kwmpi.f:.— Multiply 80 by I first, 20 X 3 = V = 15 Y = 5 x 3 = 15. COROLLARY. Every common divisor of two numbers must also divide the remainder resulting from the division of the greater of the two by the ADDITION. CA8E VI. When the numerators are alike and not more than a unit. Rule. — Multiply the numerator and denominator of the fraction having the least denominator by the common measure of the fractions. FRACTIONS. 145 Add 1 and 2, here J X 2 s» | and J make | Ans. « 1 and r « i and f " 1 and ft or thus: Add i and J, here 4- and 8 make 12 = | Ans. and 4 X 8 - 4)32 " | and & « i and f . Rule. — Add the denominators together for a new nu- merator, and multiply them together for a new denomi- nator. case vi r. When the numerators are alike and more than a unit. Rule. — Add the denominators together, and multiply thtir sum by the common numerator, and the product will be a new numerator; also, the product of the de- nominators will be a common denominator. Add 5 and ^, here 4 and 7 make 11, which multiply by the numerator 3, which is common to both. FIRST METHOD. Thus: 11 x 3 — 33 — = I/* Ans. And 4 x 7 — 28 SECOND METHOD. * x8=n f X 7 «■ H, here f | and H = H, as above. THIRD METHOD. f X f by multiplying the numer- 3 x 8 = 24 ators alternately by the denominators. 3x7 = 21 and 7x8 = || Add f andji, - - - Ans. If*. Add *and f, - - - Ans. l|f. Add i and T 7 T> ... Ans. 1 if. Add J and ft. Here the ratio between the denomi- nators is as 1 to 7; therefore, Jx7= T 'j and ft make \ \ or f answer. 13 4)1 U A iJ 5)1 5 2 5 5 2)1 12 11 1*6 FRACTIo Add i and ,V, - - - Ans. ,V Add | and A, - - - Ans. |$. Add f and / ¥ , - Ans. T 7 T . Add \ and / T , - - - Ans. ij. CASE VIII. To add mixed fractions. Rule. — Find a common denominator by reducing the fractions to the lowest terms. Add $17| " 19| " 13| " ISA $8^ 1 1 ! 1 1 ^fultiply all the divisors together 4 x 5 X 2 = 40 common denominator. CASE IX. To add mixed fractions whose numerators and denomina- tors are unlike. Add $15| Add 19| $35f (|) The operation can be performed thus, by cross multi- plication | -f- f , 24 -f- 20 = H = H reduced, from whence the following Rule is deduced: multiply each numerator by all the denominators, except its own, for a new numerator, and all the denominators together for a new denominator. Example.— Add 62$ « 37| " 19| $137H Add £, i, f , and T \ together, (say dollars.) FRACTIONS. 147 By Reduction \ is equal to 50 cents. \ « 25 « « 62* « tV " 435 " Ans. 1{| = $1.81i ad M •i H 1 Add $1 Add (4 it To add mixed or compound fractions. 1. Add J of a day f of an hour, and ^ of a minute together? Ans. 16h. 48m. 18s. 2. Add s of a year, i of a month, f of a week, J of a day, T ' r of an hour, and J of a minute together? Ans. 4m. lw. Id. 8h. 5m. 48s. 3. Add | of an eagle, f of a dollar, T \ of a dime, and i of a cent? Ans. $8.82$. 4. Add i of a week, i of a day, and £ an hour to- gether? Ans. 2d. 14*h. 5. Add | of a dollar, f of a dollar, and £ of a dime together? Ans. $1.45£. 6. Add J of a yard, ^ of a foot, and £ of an inch to- gether? Ans. 1 ft 4 in. 1 barley corn. CASE XI. To add compound fractions together, connected by the pre- position of (see Def. 9 ) GENERAL RULE. Multiply the numerators together for a new numerator, and the denominators together for a new denomina- tor. Reduce the fractions, and then add them together agreeably to Case VIII. or IX. 1. Example.— Add * of J of f, and f of 3 of \ to- gether? Ans. if. 148 FRACTIO 9 Operation, £ X f X I = -,Y T reduced is \. Now, it is plain, that | of f of f of the first compound is equal to £, and I X § X k °f ^ e second compound is equal to ^\ reduced is equal to / T > which added to | the sum is \\ as required. 2. How much is \ of \ of a dollar? Ans. 5c. 3. How much is f of T 3 T of a dollar* Ans. T » v or 18c. 4. How much is the \ of 1, the { of f , and the i of f of a dollar? Ans. 1 T ^ or $1.00c. 8*m. 5. Add i of \ of f of a dollar, to £ of 1 of f of a dollar? Ans. $/ y or 24c. Operation, $ X \ X } = ^ = T ' 5 ofa dollar or 10c And | X | X f = fa — 7 V of a dollar or 14c. Adding fractions together, #! = = 24c. 1. How much is the J and £ of i of a dollar? Ans. 50c. 2. How much is the \ and | of f of a dollar? Ans. 65c. 3. How much is 3 of | of § of a yard? Ans. 1 ft. 3 in. 4. How much is ± of \ of \ of $5.00? Ans. 12±c. 5. How much is the | and 3 of f of £ of a year? " 7m. 6. How much is the T ' v and f of T ^ of | of | of an Eagle? Ans. $1.02£. CASE XII. To reduce mixed fractions to parts, or to an improper fraction. (See 11th Definition.) Rule. — Multiply the whole number by the denomina- tor of the fraction, and add the numerator to the product for the numerator of the fraction sought, under which will be the given denominator. Example. — Reduce 17 J dollars to half dollars. ILLUSTRATION. It is well known that two half dollars are equal to one dollar; consequently, as 1 dollar is = to 2 halves, 17 units or 17 dollars will contain 17 times as much, to which if we add one-half we get 35 halves for the re- quired answer. FRACTIONS. H9 1. Bring $19 j to quarters? Ans. V $20$ to quarters? " V 33$ cts. to thirds? (i * |° 1. luring $16J to eighths? " »f 4 5. Bring $87* to halves? " "f* 6. Bring Hf to an improper fraction? " *f * TO MULTIPLY FRACTIONS, CASE I. When the fractions are proper. i.e. — Multiply the numerators together for a new numerator, and the denominators together for a new de- nominator. ILLUSTRATION. It is manifest, that when a number is multiplied by 1, the product is equal to the multiplicand; therefore, when a number is multiplied by a fraction, which is less than 1, the product must be less than the multiplicand. imple 1. — Multiply £ by £? Ans. J. From the analysis of Geometry, we find, that if a line be divided into 2 equal parts, the square of the whole line is 4 times the square of half the line: thus, let the line A — * | ! — B be one mile, yard, &c. The square of 1 is 1, because 1 X 1 is 1, and \ squared ii J, hence, £ X | mm \ of 1. CASE ir. H the multiplier and multiplicand are both mixed numbers. Rule. — Bring them to improper fractions, agreeably to Case XII. (Addition,) this done, multiply the numerators together as before, for a new numerator, and the denomi- nators together for a aew denominator; divide the new numerator (so called) by the new denominator, and the result will be the product of the mixed numbers. 13* 150 FRACTIONS. ILLUSTRATION. In the rectangular or parallelo- gram A B C D, the length of th e side A B is 10 J yards, and the length of the line AC is 1\ yards, the line A B is divided into 21 parts, riBSS8B8l»Bfi«g and the line A C into 15 equal parts, which are drawn at right angles to each other, consequently, there are 315 rectangles in the whole figure A B C D, and every four of these make 1 square yard, this is manifest from the fol- lowing example: therefore, 10 J X 7£ V X V «= *± s 78f as required. CASE III. To multiply a whole number by a fraction. Rule. — Multiply the whole number by the numerator fraction, and divide the product by the denomina- tor, the quotient will be the result. From what has been already stated, it is evident, that the multiplication of a whole number by a fraction im- plies the taking some part of it; for instance, if we mul- tiply 4 by |, agreeably to the rule 4 x | = | = 2, and 9 X i = 3,&c. Multiply 35 by *. Multiply 84 by *. « 39 by J. " 96 by *. " y J. " 80 by |. APPLICATION OF FRACTIONS TO 'SHORT ACCOl 1. Multiply Hi by 11* cts. Example, V X V = 5 i* Ans. $1.32£. 2. What will 71 lbs. come to at 8*c. pr. lb? Ans. 61 |c. :*. What will 4Ubs. come to at$| per lb? Ans. 56\c. 4. What will 19| }' ar(is come to at $§ of a dollar] Ans. $7.40§. . What will 2| yards come to at $5 of a dollar? Ans. $2.40|. li. What will 61 lbs. of tea cost at 65| cts. per pound? Ans. $4.52 5 J r . PRACTIO 151 TR ACTION OF FRACTIONS. CASE I. //' /' ■• fractions have a common denominator. Rule. — Subtract the lesser numerator from the greater, and under the remainder write the common denominator, and reduce the fraction if necessary. 1. Example — From $| take i? Ans. | of a doll, or 50c. 3—l = |=|, or .75 — 25 = .50. 2. From $5 take j}? Ans.$f or .25c. :<. From $ T V take ft} 5. From $ft 6. From I 4. From $l take J? Take $ft Take f CASS II. If the denominators of the fractions are unlike. Rule. — Find a common denominator according to Case VI. Addition, ("Second Method.") 1. Example — From \% take f? Ans. \$. Here the denominators of the fractions are in the ratio of 11 to 7, then |? x 7 = ft and |xll=H 70— 44 = ff. By Case I, or by Case VI. Addition, find a common de- nominator; thus, by cross multiplication. |? X | 10 x 7 = 70 11 x 4= || 11 x 7 = 77 common denominator, the result is \ f . 1 . From $ft take l> Ans. $ft or .02 £c. J. From $a take £? " $£ or .50c. from $|| take f ? (Here the ratio is as 4 to 1.) 4. From ${| take ftl Ans. £ or .75c. 5. From $£ take f ? " ft or .15c. 6. From $£ take ft} u ft or .45c. CASE III. When the fractions have a unit for a numerator. Rule. — Write the difference of their denominators over their product 152 FRACTIONS. 1. Example — From ^ take £? Operation, 4 — 3 __ , 4x3~ ly * 2. From $ take |? Ans. ft. 3. From £ take J? Ans. |. From ± take |? " ^ 4. From 1 take fr? Ans. J. From £ take 1? " ? \. CASE IV. When the numerators are alike and more than a unit. Rule. — Multiply the difference of the denominators by one of the numerators for a new numerator. Then mul- tiply the denominators together for a new denominator. Note. — This Rule is general, except in cases of com- pound fractions. ((^ See Case V.) Example. — From | take f . Operation, 5 — 4 = 1 x3= 3 new numerator. 5 x 4 x= 20 new. denominator. Or thus: J x 5 = |f | X4(=}| difference ft answer. From a compound fraction to take a mixed fraction. Example 1.— From } of 12, take 7f? Ans. 1±J. Operation. (According to Case XI. Addition, in rela- tion to mixed and compound fractions connected by the preposition of.) Multiply the numerators together for a new numerator, and the denominators together for a new denominator. Thus: i X V = <y ¥ = 9f Then from 9f take 7f Operation, f x 4 = if. We cannot take 15 from 12, but 15 from the common denominator 20, and 5 remains, 5 and 12 are 17; set down H and carry 1 to 7, which make 8, then 8 from 9 and 1 remains, which set down before the fraction, thus: 1£J. FRACTIONS. 153 1 From I of 8f- take 1 of 5? Ans. 1|*. From ^ of 3 take * of 2? « A- CASE VI. To subtract a proper fraction from a whole number. IE . — Subtract the numerator of the fraction from the denominator, and under the remainder place the denominator, and carry 1, to be subtracted from the minuend. Example — From $10 take § of a dollar. K) 02 $9g It is plain, that if we take I of a dollar, from a whole that § will remain. Thus, from $10.00 take 37* which'is equivalent to |,' $9.62* or g answer. DIVISION. To generalize division of fractions, the dividend must be considered as having the same relation to the quotient that the divisor has to unity, because the divisor and quo- tient are the two factors of the dividend; when for in- stance, the divisor is 5, the dividend is equal to 5 times the quotient, and consequently, this last is the fifth part of the dividend. If the divisor be a fraction, suppose |, the dividend cannot be but half the quotient, or the latter must be double of the former. The definition just given easily suggests the mode of proceeding when the divisor is a fraction. Let us take for example f , in this case the dividend ought to be only J of the quotient, but * being * of f we shall have \ of the quotient, because \ x f = iV — J*reduced by taking * of the dividend, or dividing by 4*. By having \ of the quotient, we have only to multiply it by 5, to attain_it: 1 54 FRACTIONS. thus, | x 5 = | = 1 the quotient. In this operation, the dividend is divided by 4, and multiplied by 5, which is exactly the same as taking £ of the dividend or multi- plying by |, which fraction is no other than the divisor inverted. Q. E. D. From whence, the following general rule is derived. CASE I. To divide a whole number, or a fraction by a fraction. Rule. — Multiply the whole number, or fraction by the divisor inverted. Example. Divide 9 by f . Operation. 9 is equal to f x \ = V =12. CASE II. If there be whole numbers joined to the given fractions. Rule. — Reduce them to improper fractions, and in- vert the divisor according to the general rule. ie.— Divide 9 by f ? Ans. V or 221. " 18 by |? Ans. 15. « 18 by g? . " 21f . 7± by Jl "22J. " YbyAf " 49. " tfbyH? " 1. " $14 by $|? " 35. " $22 by $|? " 55. RATIO OF MIXED NUMBERS. The following questions for exercise are well calculat- ed to exercise the learner in addition and multiplication of fractions. 1. Find 2 numbers in a given ratio, as 5 is to 6, so that their sum and product may be equal? EXAMPLE. Operation, 5 -f 6 = y = 2| and V = If; conse- quently, 2| and 1£ are in the ratio of 5 to 6. FRACTIONS. 155 Operation, 2± added to 1$ = 4 5 ' v and 2f multiplied by If — 4>J V as required; io of numbers can be proved as above, as well aa mixed numbers, whose sum and product may be equal. d 2 numbers in a given ratio, as 3 to 5, whose and product are equal? Ans. 2$ and If. 3. Find 2 numbers in the ratio of 5 to 9, whose sum and product are equal? Ans. If and 2£. 4. Find 2 numbers in a given ratio as 2 to 3, whose sum and product may be equal? Ans. 1$ and 2 J. 5. Find 2 numbers in the ratio of 3 to 8, whose sum and product are equal? Ans lg and 3^f. RULE OF THREE DIRECT IN FRACTIONS. 1. If i of a yard of velvet cost 62 -J cts. what will 2g yards cost? Operation, f X '$* X 2| = > *■&* The Rule for the common method is to invert the first term, then after preparing the fractions, to multiply the numerators together for a new numerator, and the denomi- nators for a new denominator. or thus: Let a line be drawn in all statements representing equality ■, placing multipliers on the right, and divisors on the left, and transpose the numerators when cancelled on both sides. This is a general principle in all proportional operations. 125 125 x 23 « 2875 = 7.18f. 2 2 2 4 4 V * Answer as above. 2. At 185 cts. per pound, what will 63 \ lbs. come to? Ans. $11,908. I. If i of a lb. of cinnamon bring f of a dollar, what ill If lbs. cost? Ans. $2.74*. 156 FRACTIONS* INVERSE PROPORTION BY FRACTIONS. RULE. When the fractions are prepared and the third term inverted for a divisor, as in division of fractions, then agreeably to the Rule in multiplication of fractions, mul- tiply th« numerators for a new numerator, and the denomi- nator for a new denominator. 1. How many yards of brown Holland, 5 quarters wide will line 20 yards, that is 3 quarters wide? Ans. 12 yds. 2. How many yards of matting 2 feet, 6 inches broad, will cover a room that is 27 feet long and 20 feet broad? Ans. 72 yds. 3. How much shalloon | of a yard wide, will line 4£ yards of cloth 1 \ yards wide? Ans. 9 yds. 4. What quantity of shalloon £ yard wide, will line 7 J yards of cloth 1£ yards wide? Ans. 15 yds. 5. If 3 men can do a piece of work in 4± hours, in how many hours will 10 men do the same? Ans. 1 T 7 T . 6. How many pieces of cloth at 20 J dollars per piece, are equal in value to 240f pieces at 12^ dollars per piece? Ans. 149-rY/r pieces. PROMISCUOUS EXAMPLES. 1. What will | of 2 J cwt. of chocolate come to when 6i lbs. cost | of a dollar? Ans. $10.76|$. 2. If 5 of a yard of cloth cost $2jj, what will 5/ T yards cost at the same rate? Ans. $16.25. 3. If 4i of a yard cost $9.75, what will 13£ yards cost? Ans. $29.25. 4. If $1.75 will buy 7 lbs. of loaf sugar, how much will $213.50 buy? Ans. 7 cwt. 2 qrs. 14 lbs. 5. How many yards of carpeting that is half a yard wide, will cover a room that is 30 feet long and 18 feet wide? Ans. 120 yds. * LECTURE IX. DIRECT PROPORTION. • ( Sign : : : : ) Proportion shews the direct relation of one object or thing to another, as to comparison or symmetry, viz: form, length, breadth, depth, rate, price, #c. But in rela- tion to Arithmetic or Geometry, it has a determinate meaning. Euclid has proven his Theory of proportions in the lit'th book of his Elements by demonstrating its principles, relations, and application to lines; therefore, it is apparent, that we thence derive the name of Geome- trical Proportion. In Direct Proportion, the first term is to the second as the third term is to the fourth, that is, as 2 is to 4, so is 8 to 16, written thus: 2 : 4 : : 8 : 16, these numbers are proportional, because, agreeably to the 16th proposition of Euclid, Lib. 6, the rectangle or product of the extremes is equal to the product of the means; hence, 2 x 16 = 4 x 8 = 32 or £ = y = 2. From this it is plain, that Proportion is the combination of two equal ratios, and that there are two antecedents and two con- sequents. Note to Teachers. — Require the learner to recite the answers to the following questions. Q. Why is this called the rule of Proportion? A. Because it shews the combination of two equal ratios. 14 158 DIRECT PROPORTION. Q. What is the first term of a proportion called? A. The antecedent. Q. What is the second term called? A. The consequent. Q. If 2 yards of muslin cost 25 cents, what will 6 yards cost? Ans. 75c. Q. Why is the second term (25 cts.) called the first consequent? A. Because it is the value or cost of the antecedent, Q. How are the term* arranged? A. As the first term (or antecedent) is to the second (its consequent) so is the third, (or antecedent) to the fourth, (or its consequent.) Q. How is the operation performed? A. Multiply the second and third terms together, and divide the product by the first. Q. What is the first antecedent called? A. An antecedent of the first relation. Q. What is the second antecedent called? A. An antecedent of the second relation. Q. When is a question stated correctly? A. When the first and third terms can be brought to the same denomination. Q. What is the general rule for stating? A. When a question is written thus: for instance, at 10 cts. per lb., what will 10 lbs. cost? or if 1 lb. cost 10 cts. what will 10 lbs. cost? In either case, the first term is 1 lb., then say as the first antecedent is to its conse- quent, so is the second antecedent to its consequent. Q. When the commencement of a question is written with the words, what is the value; how much, bought, or sold; what is the rule? A. In either case, the first object or thing mentioned in the question will be the third term. Q. What terms of a proportion can be contracted or reduced? DIRECT PROPORTION. 159 A. The first and third, or first and second. Q. Why except the second and third terms? A. Because the means do not shew a proper relation to each other as to cost or value; hence, it is evident, tliat tlit- second or middle term must be of the same name or denomination with the answer (or 'demand.') Q. What is meant by relation as applied to proportion? A. Nothing more than the quotient of a division. 1. If 9 lbs. of sugar cost 24 cents, what will 10 lbs. cost? Ans. 80c. lbs. cts. lbs. As 3 : 24 : : 10 = 80c. or by contraction, As 1 : 8 .• : 10 = 80c 1 lb. and 10 lbs. are the antecedents. 8 cts. and 80 cts. the consequents. The relative proportion of the antecedents and conse- quents is the same. Thus, f cts. = f $ cts. = 8. 2. If 3 yards of cloth cost $9, what will 12 yards cost? Ans. $36. 3. What is the value of 9.7 lbs. of silver at $1.5 per ounce? Ans. $174.60. 4. What will 240 bushels of wheat amount to at the rate of $6 for 5 bushels? Ans. $288. 5. How much will 17 cwt. 3 qrs. 14 lbs. of iron cost at $4.75 per cwt.? Ans. $84.90|j. 0". Sold 120 bushels of corn for $54, how much did it er bushel? Ans. 45c. 7. Bought 29 yds. of muslin for $10.87$, how much il it per yard? Ans. 37 £c. 8. Bought 2 loads of corn, one containing 75 bushels and the other 87 bushels, at 52 cents per bushel, what is the amount? Ans. $84.24. 9. Bought 3 pipes of brandy containing 120$, 124-1, and 123J gallons at 43 1 cts. per gallon, how much is the amount' Ans. $161.21 J. f 160 DIRECT PROPORTION". 10. If a staff 4 feet long cast a shade on level ground 7 feet long, what is the height of a steeple whose shade at the same time is 198 feet? Ans. 113|. 1 1 . An express who rides from Washington city had been dispatched at the rate of 60 miles per day for five days, when a second was sent to overtake him, in order to do which, he must travel 75 miles a day, in what time will he overtake the former? Ans. in 20 days. 12. A prize of $2,329 was divided between two per- sons, A. and B. whose shares therein were in proportion as 5 to 12, what was the share of each? Ans. A. $685, and B. $161 1. 13. When a man's yearly income is $949, how much is it per day? Ans. $2.60. 1 1 . Bought a stove weighing 4 cwt 3 qrs. 24 lbs. at *vMi) per cwt. and 27 lbs. of pipe at 18| cts. per lb. with 2 elbows at 50 cts. each, what is the pnce of the stovepipe and elbows? Ans. $16,481. 15. Bought 4 pieces of linen, viz: No. 1 and 2, each contained 27 i yards, No. 3 and 4, contained each 25 i yards at 62$ cts. per yard, what was the cost? Ans. $66.56i. 16. If a person's salary be $1,333 per annum, and his daily expences $2.14, how much will he save? Ans. $551.90. 17. What will 4 pieces of sattinet, containing 23, 24, 25, and 27 yards come to at 72 cents per yard? Ans. $71.28. 18. A farmer upon measuring his corn produced by a certain field, found he had but 48 bushels. It appeared that it yielded one third more than was sown, how much was that? Ans. 36. 19. A bookseller sold 10 books at a certain price, and afterwards 15 more at the same rate; now at the latter sale he received $2.25 more than at the former, what did he receive for each book? Ans. 45c. DIRECT PROPORTION. 161 CONTRACTED OPERATIONS. CASE I. If the first term be a multiple or part of the second. the third will be a multiple or part of the fourth. Example 1. — If 6" yards cost $18, how much will 24 cost? Ans. $72. CASE II. 1 1 ;i part of the first be added to or subtracted from first so as to be equal to the second, a like multiple must be added to or subtracted from the third. aple I.— If 12 yards of cloth cost $15, what will 20 yards co>t? Ans. $25. 'J. If 6 yards cost $9, what will 24 yards cost? Ans. $36. Solution. — Agreeably to Case II. the difference be- D the first and second terms is 3, and as 3, is the half of 6, A of the third term must be added to it. Thus. 24 + V = $36 as above. If 8 yards cost $12, what will 48 yards cost? Ans. $72. 4. If 45 yards cost $30, what will 165 yards cost? Ans. $110. .">. It' A yards cost $2, what will 27 yards cost? Ans. $18. <i. If 9 yards cost $6, what will 30 yards come to? Ans. $20, 7. If 3 yds cost $6, what will 10 yds. cost? " $20. mght 6 yards for $8, how much did 30 yards Ans. $40. INVERSE PROPORTI" rally admitted by writers on Arithmetic that less, or less requires more; more requires the third term is givuter than the liist, and the th less than the second. 2d. Less requires more a the third term is less than the first, and the fourth 11- H>* DIRECT PROPORTION. term greater than the second. These definitions may be illustrated by the following questions, viz: 1. If in 10 days 8 men can perform a piece of work, in what time could 40 men do the same. men. days. men. Solution.— As «8 : 10 : : 40 *= 1 x 10 1 5 5 Illustration — 8 men at work 10 days uniformly would perform as much as 80 men in 1 day. Consequently, if 80 men perform the same in 1 day, 40 men will do it in 2 days. Q. E. D. 2. If 8 men can perform a piece of work in 10 days, how many men would be required to perform the same in 2 days? Ans. 40. days. men. days. Solution— As 10 : 8 : : 2 : 40. Notwithstanding, these statements are made to corres- pond with the above definitions, and bring correct an- swers. Yet, it is manifest, they are wrong ! ! ! if the doctrine of proportion be recognized, for agreeably, to Prop. B. book 5 Euclid, in Inverted Proportion, the third is to the second, as the first is to the fourth. In solution 1st, we find that more requires less for 10 days require 2 days, and in solution 2d, less requires more — for 8 men require 40 men. Now the proper method of stating is thus: as 40 men, : 10 days : : 8 men = 2 days. Solution 2d — 2 days : 8 men : : 10 days = 40 men. Hence, 40 x 2 = 10 X 8, And 2x40= 8 x 10— for by Prop. 16, 6 Eu- clid, the .rectangle of the extremes will be equal to that of the means, which is not the case with the first and second statements; therefore, it follows, that although custom and the opinion of most writers, sanction the truth of the l first and second' statements they are evi- dently mistaken. DIRECT PROPORTION. 163 3. Suppose 450 soldiers are in a garrison, and their provisions are calculated to last them but 5 months, how many must leave the garrison, that the same allowance may be sufficient for those who remain 9 months. Ans. 200. 4-. If a man perform a piece of work in 15 days of 12 hours long, in how many days of 10 hours long can he perform the same? Ans. 18 days. re is a cistern having a pipe that will empty it in (J hours, how many pipes of the same capacity will empty it in 20 minutes? Ans. 18 pipes. COMPOUND PROPORTION. Is that, in which 5 terms are given to find the sixth. Three of those terms are a supposition and two a demand. Rule. Arrange the terms so that two terms, one side of the statement shall be of the same name and denomi- nation with two terms on the other side. Consequently, the demand will be in the middle term. ample. — If 10 men in 18 days earn $56, how many dolled can 20 men earn in 36 days? men. demand. men. Mode of operation 10 ) $56 (» °20 2 by contraction days 18J 4. ^ "36 2 4 1)224 $224.00 Answer. 1 1*7 reapers get $21 for 3 days work, how many ri will earn $96 in 32 days? Ans. 3 reapers. :*. If a family of 8 persons in 24 months spend $480, how much would 16 persons spend in 8 months. Ans. $320 00. \. [f 4 men mow 96 acres in 12 days, how many acres 8 men mow in 16 days? Ans. 256 acres. 1 f 1 men receive $24 for 6 days work, how much will S men receive for 12 days work. Ans. $96. PRACTICAL ARITHMETIC. AS ABBREVIATED IX THE COUNTING ROOM, OF CURRENCIES AND COINS. u cent*. f. d. u «. d. J 6d. $ 1 5 -- 50 i ^ 10.0 J ^ 1.8 T ' r 5 * £ I 25 \ a 5.0 4 i-o^ ^ 3 4 ^ 20 | g. 6.8 4 0.G ,', - 2 A = ;- 124 \ 4.0 t FJ °.*A i ,v 1 1 64 A 10 T ' r 3.4 J 2.6 J 1 °- 3 ^ OF WEIGHTS. \\i loan. WllGHTS. Wkk;htb. Parts of a Ton. Par/* of a at*. Parts of a cwt. WVijA/*. 20 CWt 1 ton 4qrs. 1 cwt To take 161b. | 1 1 •• J 10 * " 2 " J " of 3 5 4 « 1 " i " qrsof a cwt. 8"tV 4 I " 1 28 lbs. 1 qr. 2qrs. i 2 T ^« 14 « 4 « 1 « 1 1 A" 7 « I * of "J |L_ OF MEAS1 i:es. Jtfeo«vrM. Measures. Measures. ires. Measures. Paru of a yd Pan» of a bu. Pans iff a gal. Purls of an ac Parts «f an acre. 4 qr 1 yd 4 p. 1 bu. 4 qts. 1 gl 160 ps la. 4 rds. 1 a. 2 " i " 2 u . u 2 .. j u 80 " 4 " 2 " I " 1 " 4 " 1 « 4 " 1 « I * 40 " 4 <k 1 « 4 « nails iMarts. pecks pmts. 20 " 4 " parches, 4 4 of yd 8 1 2 1 qt. 10 "-r 1 *" 20 4 rood. 2 4 w 4 1 I •• 5 «,V< 32 m £ « 10 4 " ItV " 2 4 1 1 1 I * 8 | " pts.ofgal. 2 - 1 PRACTICAL ARITHMETIC. 1 08 CASE I. 1 low to take parts for | of a cent, | of a yard, | of a I of a cwt. f of a year, &c. Hi i.e. — Take the aliquot parts for | first, and then for cl add the product together; or, multiply the given quantity by 3, and divide by 4. If it be \, divide by 4, . divide by J CASK II. ii the given price is dollars and cents. Ri le. — Multiply the given quantity by the dollars, and take aliquot parts of the cents, and add the products •her. CASE III. When the given quantities are of various denomina- tions, as cwts. qrs. and lbs.; yards, quarters, nails; or bushels, pecks, and quarts, &c. Rule. — Set down the price per cwt., multiply it by the number of cwts. given, and take aliquot parts, for the quarters and lbs. &.c. Q. What is the amount of 15 cwt., 2 qrs, 7 lbs. of iron at $10 per cwt? Operation $10.00 X 15 $150.00 2 qrs. = ^ cwt. ) 7 50 price of f a cwt. 7 lbs. = i of 2 qrs. J 0.93£ price of 7 lbs. $158.43£ 1 . What will 36 yards of tape come to at ± of a ct. per yd. •J. M « 96 a % M "295 ii 4. M ii 5. M "440 ii H "896 ii * i it " i it ? 1 it ii f H * 1 tt 166 PRACTIEAL ARITHMETIC, CLOTH MEASURE. yd*, yr*. | di. 1. What is the value of 27 3 at $9.65 pei yd 1. 267.79 yd*, qr. nls. 2. u 860 1 at 84 cts. $722.61 3. « 126 2 2 « $4.75 601.17 4. u 7 3 2 - ; 5.60 44.10 r>. 4 29 1 3 " 3.75 DRY MEASURE. bush. pks. 110.39 1. What is the v bush. alue of 120 2 at 35 cts. per pks. qts. pts. 42.17J 2. « 780 8 v> at $1.17 per bush. 913.55 1 « 1354 15 1" 25 * 338.601 4. (4 325 3 « " " 62i « 203.59| 5. U 94 2 " " " 87i « 82.68f LIQUID MEASURE. 1. What will 784 gals come to at 84 cts per gal. $658.56 gals, qts. pts. Ans. $ cts. 2. " 765 3 1 at $2.18| per gal. 1675.35 3. « 5 hhds. 31 i gals, at $47.12 per hhd. 259.16 4. " 17 " 155 or 3 qts. 63.75 M 1116.93| 5. " 428 gals. 3 qts. at 1.40 LAND MEASURE. 600.25 A. R. P. 9 <**. 1. What is the value of 35 2 9 at $54.35 1935.54 A. R. P. 2. « 146 3 10 at $35.10 5153 12 3. * 750 1 "4" \>:.-^ 9190.87 4. »< 17.) 3 12" 52.15 9160.274 :>. '• 196 2 32" 41,10 8674.47 6. M 189 3 395 4-9.991 9999.42 EXERCISES FOR THE SLATE. Tons. act. qr. lbs. 19 19 3 27£ at $99.99| per ton, Ans, $1999.93 Xow 20 tons at $100 per ton is $2000.00 PRACTICAL ARITHMETIC. 167 Deduct, 20 tons it £ c $0.05 Again, we find that , of a ton 4480) 99.99 75(=0.02 ) 0.07 As before $1999.93 t will 29 J yards come to at $4- per yd. Operation 30 yds. — J =29 J 30 yds. at $4 ==$120.00 Deduct i of $4.00 m 0.50 $119.50 yds. 30$ at $2.40 per Ans. $74.00 19? « 1.75 « « 34.56i i. m 44| « o.80 M « 35.50 5, « 12f " 3.15 " " 39.77 6. « 26* " 3.37* « « 88.59* A merchant in Wilmington (Del.) bought the follow- ing bill of goods in Philadelphia for cash, and calculated to dispose of them, at 25 per cent profit. jFtr«* cost. 1 pieces Irish linen 236 yds. at 1 bale shirting muslin 400 * 19 pair shoes 12 «* boots 50 yards white flannel 50 « ticking 120 ' 4 bombazine 44 cts. $103.84 15 " 60.00 1.75 " 33.25 2.25 " 27.00 44 " 22.00 20 « 10.00 45 « 54:00 $310.09 To gain 25 per cent, add i 77.52J To mark goods, assume some word containing 10 let- ters; suppose the word Perthamboy Let these ten letters be represented by each of the nine 168 1>RACTICAL ARITHMETIC. digits, with a cipher, viz: in order to obtain private marks for the retail price of the goods. — Thus, The In.-h linen was purchased for 44 cts. To gain 25 per cent, add i 11 55 retail price. Hence the private mark is hh. In like manner, the shirting cost 15 cts. To gain 25 per cent., add \ 3| 18| retail price. Consequently it can be marked pb r-t. Again, the shoes cost $1,75 per pair. To gain 25 per cent, add 0.43| $2.1 8| selling price. e-pb r-t, private mark. The boots cost $2.25 per pair. To gain 25 per cent add 0.5b"i $2.8U selling price. The flannel cost To gain 25 per cent, add \ e.bp p-t, private mark, per yard. 11 The ticking cost To gain 25 per cent add i 55 selling price. hh private mark. 20 cts. per yard. 25 selling price. The bombazine cost Add \ to gain 25 per cent. eh private mark. 45 cts. per yard. 56^ selling price. ha.p-t. private mark. PRACTICAL ARITHMETIC. 169 PROOF. yds. marked. 10 pieces Irish linen 236 at55c( hh ) $129.80 1 bale shirting muslin 400 at 18| ( pb r-t) 75.00 I ! » pair shoes at $2,181 (e.pb r-t) 41.56* boots at 2.81* (e.bpp-t) 33.75 50 yards ticking 0.25 ( eh ) 12.50 50 " flannel 0.55 ( hh ) 27.50 1'20 " bombazine 0.56* ( hap-t) 67.50 As before, $387.61* 1. Bought 260 yards double milled cassimere at $3.00 per yd. Required the private mark at a gain of 20 per cent, profit Ans. $936.00 Selling price $3.60, private mark u r.ay. il J. Bought 50 pieces domestic muslin, each 29 yards, at auction, at 10 cents per yard, which I disposed of im- mediately at 10 per cent profit, what is the "private mark" and the selling price? Ans. lie. And the private mark pp. CASE III. 1. What is the price of 5 cwt. 1 qr. 14 lbs. at $2.50 Ans. $13.43|. lbs. 7 at $10.25 per cwt. Ans. $108,261. 19 at 4.15 " « 32.86*. Oat 5.18 " " 85 47. Hat 27.10 " " 16.93*. cents per yard, what will 60 yard* three- quarters come to? Ans. $41.91 1. 2. At $1.46 per yard, what will 33 yards three-quar- ters come to? Ans. $49.27*. 3. At 43| cents per yard, what will 81 yards three- quarters come to? Ans. $35.77. 4. At $1.64 per yard, what will 104 yards, three-quar- ters, and 3 nails come to? Ans. $172.09|. 15 ;r cwt.? cwt. 2. 10 qrt. 2 3. 7 3 4. 16 2 5. 2 1. At 69 170 PRACTICAL ARITHMETIC. 5. At 33 3 cents per yard, what will 63 yards of inus~ lin come to? Ans. $21.00. 6. At 31 i cents per yard, what will 40 yards come to? Ans. $12.50. 7. At 93| cents per yard, what will 80 yards come to? Ans. $75.00. 8. What will 3 yards, 3 quarters and 3 nails of cassinet come to at } of a dollar per yard? Ans $2.46. 9. What will 65 yards, 2 quarters, and 1 nail come to at I of a dollar? Ans. $57.3* 10. What will 420 yards come to at T 8 ff of a dollar? Ans. $?v 1 1 . What will 35 yards come to at T 9 f of a dollar? Ans. $19.Cs 12. What will 42 yards come to at -,\ of a dollar? Ans. $18,371. 18. \\ ! .t will 63 yards and 3 quarters come to . of a dollar? Ans. $19.92*. 14. Bought 2001bs. cheese at 9fc. pr. lb. " $18.: 15. Bought 12 bushels, 2 pecks, 4 quarts of apples at 33 i cents per bushel? Ans. $4.21. Bought 12 gallons, 3 quarts, 1 pint of brandy at $1.05 per gallon? Ans. $13. J. 17. Bought 12 lbs. 14 ozs. of tea, at $1.37£ cents per pound? Ans. $17.70. 18. Bought 25 lbs. 12 ozs. of nutmegs at $3.25 cents per pound? Ans. $83.68|. CUSTOM HOUSE CALCULATIONS. To secure the exact collection of duties imposed on certain articles of imported merchandize, and on the ton- nage of vessels employed in commerce, the law provides that the cargoes of vessels employed, in foreign commerce shall be inspected, and weighed or guaged by the custom house officers, and certain allowances made on boxes, casks, &c. containing goods on account of leakage, break- age, &c. PRACTICAL ARITHMETIC. 171 ALLOWANCES. Draft, is an allowance, made from the allowance of !c, &c. on account of probable waste. is an allowance made for the weight of the box, bag, cask, &c. containing the goods. Gross ux it: fit is the whole weight, including the weight of the box, bag, or cask, containing the goods. JVW u-f ighi is the weight of any parcel of goods after the draft and tare have been deducted. DRAFT. On a single box, &c. weighing 1 cwt. or 112 lbs. 1 lb. On weighing above 1 cwt. and under 2 cwt. 2 lbs. " « " 2 cwt. and under 3 cwt. 3 lbs. " " " 3 cwt. and under 10 cwt. 4 lbau " " "10 cwt. and under 18 cwt. 7 lbs. " " * 18 cwt. and upwards, 9 lbs. TARE. Tare is computed on the remainder of any weight af- the draft has been allowed. On sugar in casks (except loaf) is 12 per cent. On M in boxes, - - 15 4< On " in bags or mats, - .5 " On cotton in bales, - 2 " On " in ceroons, - 6 li On Glauber salts in casks, 8 " On nails in casks, - - 3 a On pepper " - - 12 " On « in bales, - - 5 « On " in bags, - - 2 " On sugar candy in boxes, - 10 " On twine in casks, - - 12 " On " bales, - 3 « On cheese in hampers or baskets, 10 (l On " in boxes, - - 20 " pn candles " - 8 * Ill PRACTICAL ARITHMETIC. On soap in boxes, - - 10 per cent, On chocolate " 10 " On shot in casks, - 3 u T'/rc on all other goods paying a specific duty is allowed according to the statement of the same in the invoice, which is considered the actual weight of the box, bag, cask, &c. The importer may have the invoice tare allowed, if he make his election at the time of making his entry, with the consent of the Collector and Naval officer. Leakage allowed on a.11 merchandize in casks on the guage, paying duty by the gallon, is 2 per cent. Breakage, 10 per cent is allowed on all beer, ale, and porter in bottles, and 5 per cent, on all other liquors, or the importer may have the duties computed on the actual quantity at the time of entry. Common size bottles are estimated at the Customhouse to contain 2£ gals. pr. doz. Duties on foreign goods, are either ad valorem or specific. Ad valorem duty is a certain per cent of the actual cost of the goods in the country from which they are brought. Specific duty is fixed at a certain sum per ton, hundred weight, pound, square yard, &c. Tare, draft, #c.'are to be made, before the duties are computed. 1. Calculate the duty on an invoice of dry goods which cost in Liverpool $9,840 at 10 per ct ad valorem? Ans. $984.00. ( alculate the duty on 3 pipes of wine, allowance for leakage as in the table; duty 7| cents per gallon? Ans. $27 78. 3. Calculate the duty on 10 gross of London porter, allowance for breakage as in the table; duty 20 cts. per gallon? Ans. $57.60. 4. Calculate the duty on an invoice of silk goods, which cost in Canton $6,400 at 10 per cent ad valorem? Ans. $640. TARE AND TRET. 173 5. Calculate the duty on 4 casks of Rochelle salts, in- voiced at $10 per cwt. gross weight as follows: 1st cask, 1 cwt 2 qrs. 12 lbs.; 2d, 1 cwt. 1 qr. 17 lbs.; 3d, 2 cwt. 3 qrs. 7 lbs ; 4th, 4 cwt. 1 qr., draft as in table; tare 8 per cent, duty 15 per cent ad valorem? Ans. $14.09- TARE AND TRET. d is an allowance made on some particular articles on account of waste. Tare is an allowance made by the seller to the buyer for the weight of the case, cask, box, bale, &c, in which goods are packed, calculated, at so much per cask, or at so much per cent, according to the nature of the goods. Neat weight, is 1toat quantity to be settled for, after all Allowances have been deducted. CASE I. When the Tare is so much 'on a given quantity gross. Rule. — Subtract the given tare from the given quan- tity, and the remainder will be the neat weight. 1. What is the neat weight of 1 hhd. of tobacco weigh- ing 5 cwt. 2 qrs. 15 lbs. gross, when the tare is 3 qrs. 7 lb? Gross, 5 2 15 Tare, 3 7 Neat wt. 4 3 8 Answer. 2. What is the neat weight of 8 hhds. of sugar, each weighing 7 cwt. 3 qrs. 20 lbs., tare in the whole 5 cwt. 1 qr. 19 lbs. Ans. 58 cwt. qrs. 1 lb. CASE II. i the tare is so much per bag, box, hogshead, or other denomination. Rule. — Multiply the given tare per bag, box, barrel, &.c. by the number of bags, boxes, barrels, Sec. and sub- tract the product from the gross, the remainder will b* ihe neat weight. 15' 174 TARE AND TRET. 1. How much is the neat weight of 25 kegs of raisins, each, gross 1 cwt. 2 grs. 15 lbs., tare 19 lbs. per hundred weight. act. qr». lb*. 1 2 15 5 X 5 = 25. 8 19 5 Gross weight, 40 3 11 16Jbs. is* = 5 2 lbs. is J 1 lbs. is i 19 =» Tare, 6 3 18 Neat weight— cwt. 33 3 21 Answer. 2, What in the neat weight and cost of 10 hhds. of tobacco, each weighing 5 cwt 1 qr. 13 lbs. gross, and 16 lbs. tare per cwt. at $8.75 per. cwt.? Ans. neat 46 cwt. cost $402.50. CA8E III. When the tare is so much on the hundred toeight. Rule. — Divide the gross weight by the aliquot parts of a hundred weight, and deduct the amount of the result from the gross, and the remainder will be the neat weight. Case III. will be found applicable to the last examples. CASE IT. When the Tare and Tret are both allowed. Rule. — First find the tare, which deduct from the gross, and the remainder will be the 'suttlej divide the suttle by 26,* the product will be the trett, which sub- tiact from the suttle, the remainder will be the neat weight. * Four pounds on the 104 lbs. is the usual allowance for tret; the reason we divide by 26 is, that 4 lbs. is 1-26 of 104, TARE AND TRET. 175 1. There are 17 boxes of sugar, each 1 cwt. 3 qrs. 18 lbs. gross, tare 16 lb. per cwt., trett 4 lbs. per 104 lbs.; what is the neat weight, and what is the value at $7.60 pr. cwt? cwt. qrs. lbs. 1 3 18 X 1 $7.60 qrs. lbs. 4x4 = 16 26 3 4 4560 1520 7 2 16 4 30 1 » 2 3 8 18 16 lbs. | \ | 32 4 1 2 26 15 27 Tret, 41b. & 1 3 11 7 or. 2 1 1 I lb. 4 t 19760 380 190 027 -f $203.57 26 3 4 neat. 2. What is the neat weight and value of 12 bags of coffee, each 2 cwt. 1 qr. 10 lbs. gross, tare 18 lbs. per cwt., tret 4 lb. per 104 lbs. at $19.60 per cwt.? Ans. 22 cwt. 2 qrs. 18 lbs. cost $444.15. When the tare is rated at so much per cent on the gross. Rule. — Multiply the gross by the tare per cwt. and divide by 100 for the tare, then calculate the amount as before. 1. What is the value of 15 hhds. of loaf sugar gross weight 68 cwt. 3 qrs. 21 lbs., tare 12 per cent, at 14 cts. per pound? Ans. $951.22$. J. What is the value of 4 hhds. of tobacco, weighing as follows: Tare L2 lb. per 112 lbs. at $3 75 per cent. Ans. $93.78. cuts. qrs. lbs No. 1 r= 6 3 18 "2 = 7 10 " 3 — 5 3 26 « 4 — s 03 Gross weight, 28 01 176 INTEREST. INTEREST. Q. What is the consideration or legal allowance for the use of money called? A. It is called interest or premium. Q. What is the sum lent called.' A. The principal. Q. What is the legal interest or premium of $100 per annum called? A. The rate per cent. Q. Why is it called the rate per cent? A. Because, for a loan of $100, six dollars is allowed for its use for one year, it being considered the legal per centage in every State in the Union, except New York, in that state, the rate is $7 per annum. Time. — The number of years, months, or days for which interest is to be computed. Jlmount. — The principal and interest added together is called the amount GENERAL RULES AND THEOREMS FOR CALCULATING SLM PLE III rEREST IN DOLLARS AND CENTS. THEOREM L This theorem is deduced from the principle, that any sum, at simple interest will double itself in 16 years, 8 months, or 200 months. Putting P = any given principal, T = any given time in months we have, T P TP 200 : T : : P : — X m which expressed in 2 100 200 words, reads thus: Rule. — Multiply the given principal by half the giveng *ime in months, pointing off two figures on the right hand for decimals. IEREST. 177 THEOREM II. rem furnishes the shortest method possible for calculating simple interest for days: it is well known, hi hanking institutions, interest is universally charg- ed on all loans at the rate of one per cent, for sixty days, iefore, it is manifest, that putting P = any given principle, and T = any given t'ine in days less than a T year, we shall have 60 : T : : 1 = e is = the interest of $100 for the given time. And again, T T x P T P 100 : P : : — : = — X = the 60 6000 6 1000 interest of P for the whole time. Rule. — Multiply the principal by one-sixth of the n number of days, and divide their product by 1000, for cents, or point off 3 decimal places on the right hand, and you have the interest required. To find the interest of any sum for days. Rule. — Multiply the principal by & of the days which denote as so many thousandths for decimals, the pro- duct when pointed will be the interest. By analysing, we find that as $100 gives $6 interest in 860 days, it will give a proportional interest for any other number of days, of which 360 is a multiple. Therefore, it will be clearly seen, that the interest of $100 for 60 days at 6 per cent, is $1.00; hence, it is manifest, that the interest for any given sum for 60 days, will be as many cents as there are dollars in the principal, for instance, the interest of 48 dollars for 60 days, is 48 cts. s50 principal interest, 50 cents, and so on. 1. What is the interest of $80 for 60 days? Ans. 80c. h interest of $90 for 60 days > " 90c. 3. What is the interest of $40 for 60 days? " 40c. 4. What is the interest of $45 for 60 'days? " 45c. 178 INTEREST. Equimultiples of sixty days. From the preceding illustration it is plain, that if the interest of 80 dollars for 60 days be 80 cents, the interest for 30 days will be 40 cents. 1. What is the interest of $240 for 120 days? Ans, $4.80. / -hat ion. — The interest of $240 for 60 days would be 240 cents or $2.40, and as the ratio of 60 to 120 is as 1 to 2. Consequently, $2.40 x 2 = $4 80. 2. What is the interest of $300 for 120 days? Ans. $6.00. 3. What is the interest of 350 for 120 days? " 7.00. 4. What is the interest of 460 for 120 days? « 9.20. METHOD OF CALCULATION IN HANKING INSTITl TK>N> Rule. — .Multiply the principal by the number of days, including the days of grace, and the day on which the money was paid to the drawer, and divide by 60. 1. What is the interest (or baak discount,) on a note of $100 for 60 days. 6400 .60 + 4 - 64 x 100 =* c= 1062 or $1.07. 60 But the usual method of Bank calculation is more con- cise; thus, $100 for 60 days, agreeably to the analysis of Theorem 2, is $1.00. 3 days grace -f 1 day for payment. 4 days being = to T ^ of 60 is equal to .07 nearly, $1.07 as before. 2. What is the interest on a note of $100 for 90 days? Ans. $1.57. Interest of $100 for 60 days is - - $1.00 100 for 30 days is - - 0.50 Interest for 3 days grace and day of payment, ^ viz; 4 days is - - .07 $1,57 INTEREST. 179 3. interest on a note of $100 for 116 days? Ads. $2.00. Operation.— 116 + 4< = 120 = 60 X 2. The interest of $100 for 60 days is - $1.00 Itiply by 2 Answer, $2.00 ANOTHER METHOD. Rule. — From the amount of the note, point of two figures to the right hand, this done, divide by 15, add the quotient to the principal, and the sum will be the interest required. I . Calculate the interest on a note of $100 for 60 days? 1.00 Illustration. — .062, which added to $1.00 = 15 $1.00 + .061 = $1,061 in Bank, $1.07. 1 f for 90 days analyse it thus: for 60 ds. set down $1.07 for 30 ds. » 50 sl.fw Find the Bank interest on a note of $100 for 30 days? Ans. $0.57. 6. Find the Bank interest on a note of $240 for 90 days? Ans. $3.67. Note. — When no per centage is named, 6 per cent, is tood. CASE II. ■t the time is one year, and the rate per cent, any num- ber of dollars. Rule. — If the principal be dollars, point out 2 places lor cents. 2d. If it be in dollars and cents, point out 4< places for the decimal parts of a dollar. 3d. If in dol- cents and mills, point out 5 places for the correct er. 180 INTEREST. 1. What is the amount of $144 for 1 year at 6 per cent? Ans. $152.64. 114 6 Interest, $8.64 We point out 2 places for cents Principal, 14-4.00 agreeable to Rule 1st. Amount, $152.64 2. What is the amount of $240.75 for one year at 6 per cent? Ans. $255.1 9 J. 3. What is the interest of $640 for 1 year? Ans. $38.40. 4. What is the interest of $20.33, " « 1 42. 5. What is the interest of $1620 '« « 97.20, 6. What is the interest of $19.64 for 2 years] Ans. $2.36 nearly. To find the interest for years and months. CASE III. Bring the years to months, and multiply the principal by the number of months, and half the product will be the interest, or take half the months and multiply it by the principal. 1. What is the interest of $325 for 12 years? Ans. $234.00. 2. What is the interest of $617.56 for 25 years? Ans. $926.34. 3. What is the interest of $17,696.56 for 10 years? Ans. $10,617.93. 4. What is the interest of $250 for 1 1 years? Ans. $165.00. 5. What is the interest of $13.93| for 3 years? Ans. $2.50. To find the interest for months at six per cent. Rule. — Multiply the principal by half the number offl months, and proceed as before. 1. Find the interest of $240.75 for 2 months at 6 per cent per annum? Ans. $2.40|. INTEREST. 181 interest of $320/25 for 3 months? Ans. $4. 80.3 + Aterett of $480.90 for 4 months? Ans. $9.61.8. !. lui.l the interest of $325.92 for 5 months? Ans. $8.14.8. 5. What is the amount of $240.92 for 2 years, 10 months at 7 per cent.? Ans. $288.70.2. 6. What is the amount of $325.15 for 3 years, 5 months, at 5 per m Ans. $380.70. CASE IV. When the rate of Interest is at any per centage more or less than six per cent. Rule. — Find the Interest ot the given principal at 6 per cent., as before directed; then multiply that interest by the given rate per cent, more or less than 6 per cent, and divide the product by 6, and you get the interest required. 1 . Calculate the interest on a note of $275 for 4 months at 7 per cent.? Ans. $6.11$. 2. Calculate the interest of $350 for 5 months, at 5 per cent.? Ans. $7. 29 -f 3. Calculate the interest of $248.75 for 1 1 months, at 7 per cent.? Ans. $1.\96. 1 . Calculate the interest of $365.67 for 7 months, at 3 per cent.? Ans. $6.40. 5. Calculate the interest of $600 for 15 months at 5 per cent Ans. $37.50. Calculate the interest of $9Q.80 for 19 months, at 6 at per annum? Ans. $8.62.6. CASE V. When the Interest is required for any number of years and months. Rule. — Bring the years and months, to months, take half the sum; multiply that half sum, or half the number 16 1S2T INTEREST. of months, by the principal, for v the interest at 6 per cent, according to Rule, 2nd case. 1. What is the interest of $65 for 3 years, 4 months, at 6 per cent, per annum? Ans. $13.00. 2. What is the interest of $199.11 for 3 years 8 months at 6 per cent? Ans. $43.80.4.. 3. What is the interest of $98 for 4 years 2 months? Ans. $24.50. 4. What is the interest of $1298,40 for 4 years 8 months? Ans. $363.55.2. CASE vi. To find the Interest of any given principal for any num- ber of years, months and days, at six per ct.per annum. Rule. — Bring the years and months to months, take one-sixth of the days, which annex to half the months, multiply that number by the principal, and point out three decimal places in the product. One for mills and two for cents, which will give the correct answer. It may be asked, why we take \ of the days, an ai. is at hand; because, 30 days is the general average of a month; — Days are therefore; thirtieths of a month, and sixtieths of half months, because the Interest of any sum at 6 per cent, is exactly $ per cent, a month. — Hence if \ of the days be taken they will become tenths, or deci- mals of a month. 1 . W T hat is the interest of $400 for 3 years, 4 months and 12 days, at 6 per cent, per annum? $80.80. yrt. m. d. llustration, 3 4 |)12 • -2 i *)40 202 400 $80.80.0 Answer. st. 183 interest of $360 for 2 years 8 months Ans. $58.68. llate the interest of $800 for 1 year 8 months and 6 da\ Ans. $80.80. I. Calculate the interest of $240.65 for 4 years, 4 months and 15 days? Ans. $63.17. Calculate the interest of $67.50 for 1 year, 7 months and 7 d $6.49 Calculate the interest of $23.19 for 2 years, 5 months and lid. $3.1 CASE VII. When the Interest is required from a certain day of ihe month, in year, to a particular day of the month in th> or in a not fur year. 1. What is the interest of $30, from Feb. 6th, 1838, up to June 22d, 1840? Ans. $4.28. 'J. What ia the interest of $35.61 from Nov. 11th, 1831, Dec. ! >th 1833? Ans. $4.47.4. 3. What is the interest of $11.10$, from April 17th, to . 7th, 1832, at 7 per cent.? Ans. $0.49.6. 1 . What is the interest of $369.29, from April 30th 1830, to July 31st 1832, at 4 per cent.? Ans. $33.28. W t is .the interest of $1728.75, from Nov. 19, . up to June 18th, 1826? Ans. $267.66.8. I is the interest of $99.99.9, from Jan. 1st, 1800, to Feb'y 29th, 1832? Ans. $192.96.4. PARTIAL PAYMENTS ON BILLS AT INTEREST. In the Supreme Court of the United States, and, in- deed, in most courts of the several States, the following rule is generally adopted, for estimating interest on Notes and Bonds; when partial payments have been made. GENERAL RULE. Compute the interest on the principal sum, from th,e when the interest commenced, to the first time when I made, add that interest to the principal, 184 WW and from the sum subtract the payment made at that time, together with the preceding payments, if any, and the re- mainder forms a new principal; on which compute and vibtract the interest, as upon the first principal, and pro- ceed in the same manner to the time of judgment. Wilmington, Del. June 17, 1829. For value received, I promise to pay Joseph Miller, or order, on demand, seven hundred and sixty-nine dollars, and eighty-seven cents, with interest. $769.87. NELSON CLELAND. Attest, Eli Hillis. On this note are the following payments: March 1 , 1830, received seventy-five dollars and fifty cents. June 11, 1831, received one hundred and sixty-five dollars. September 15, 1831, received one hundred and seventy- one dollars. January 21, 1832, received forty-seven dol- lars and twenty-five cents. March 5, 1833, received twelve dollars and seventeen cents. December 6, 1833, received ninety-eight dollars. July 7, 1834, received one hundred and sixty-nine dollars. What remains due Sept Ans. $211.92. METHOD OF OPERATION. Principal carrying interest from the date of the note, June 17, 1829, to March 1, 1830. yrs. m. d 1830 3 n Principal, $769.87 1829 6 17 I Int. for 8m. 15d. 32.72 diff. of time. 8 15 J $802.59 Subtract first payment, 75.50 Balance for new principal, $727.09 yrs. m. d. 1831 6 11 1830 3 1 Interest, 55.74 diff. of time. 1 3 10 782.83 Subtract second payment, 165.00 Balance for new principal, $617.83 t I EREST. 185 yr». m. d. isji 9 15 1831 6 11 Int. for 3m. 4d. 9.68 3 4 $627.51 Subtract third payment, 171.00 Balance for new principal, $456.51 yr*. to. d. 1831 9 15 Int. for 3m. 6d. 7.30 t time. 3 6 $463.81 Subtract fourth payment, 47.25 Balance for new principal, $416.56 yrs. to. d. 1834 3 5 $416.56 1832 1 21 Int. forly. Im.l5d. 28.12 diff. of time- 1 1 15 $444.68 Subtract fifth payment, 12.17 Balance for new principal, $432.51 yrs. to. d. 1833 12 6 1833 3 5 Int. for 9m. Id. 19:54 ldi£ of time. 9 1 $452.05 Subtract sixth payment, 98.00 Balance ibr new principal, $33 1 i )5 yrs. to. d. 1 S34 7 7 1833 12 6 Int. for Oy. 7m. Id. 12.45 ilill'. of time. 7 1 $366.50 Subtract seventh payment, 169.00 Balance for new principal, $197.50 yrs. to. d. 15 9 25 1834 7 7 iff. of time. 1 2 18 Int.for ly.2m. 18d. — 14.42 Due on the 25th September, 1835, $211.95 16* 186 INTEREST. Baltimore, March 10, 184-0. For value received I promise to pay Simon Kemp, fifty four dollars and eighteen cents, with interest. $54.18. - JAMES HOOPER. Attest, John E. Stansbdrt. On this note are the following payments: June 10, 1840, received twelve dollars and twenty-five cents. August 16, 1840, received ten dollars. October 21, 1840, received twenty dollars. March 4, 1841, received five dollars. What sum will be due on the 4th day of July, I841J Ans. $9.04. MISCELLANEOUS QUESTIONS IN INTEREST. CASK I. Principal interest and time given to find the rate per cent, 1. At what rate percent, must $500 be put on interest to gain $ 1 20 in 4 years? Rule. — Multiply the interest by the time, and subtract the product from the amount, the remainder will be the interest gained at 1 per cent, for the given time, provided the difference be less than the principal, divide the inter- est by that remainder, and the quotient will be the rate Mf pent n quired. ///usfra/ion—120 x 4 = 480, 500 — 480 = 20, and '.%• = 6 per cent as required. Proof. — $500 at 6 per cent per annum gives $30 in- terest, which being multiplied by 4, = $120 the interest of $500 for 4 years. & If 1 receive $60 interest for the use of $600, 1 year, 8 months, what is the rate per cent. Ans. 6 pr. ct. 3. If I pay $200 for the use of $2000, for 2 years 6 months, what is the rate per cent? Ans. 4 per ct. 4. At what rate per cent, must $400 be put to interest to gain $120 in 5 years? Ans. 6 per ct. j 5. At what rate per cent, must $500 be put on inter- est to gain $120, in 4 years? Ans. 6 per ct. INTEREST. 187 CASE II. I interest multiplied by the time exceeds the principal. 1. At what rate per cent, must $125, be put to interest to gain $37.50 in 6 years. Ans. 5 per ct. J. At what rate percent, will $480 yield, $90 interest in 3 years, 1 month, and 15 days? Ans. 6 per ct. 3. If $ 225 gain $ 1 08 in 8 years, what is the rate per Ans. 6 per cent. 4. At what rate per cent, must $120 be on interest to amount to $133.20 in 16 months. Ans. S\ per cent At what rate per cent, must $280 be on interest to amount to $411.95 in 6£ years? Ans. 1\ per cent. CASE II. 1. Suppose $1000, at 4£ percent per annum, amount to $1281.25. How long was it at inU; Ans. 6 years 3 mos. J. In what time will $1600 amount to $2048 at 4 per cent, per annum? Ans. 7 years. case in. 1. In what time will the interest of $600 be equal to the principal at 6 per cent.* Ans. 16 years 8 mos. GENERAL RULE. Divide 100 by the given per centage, and the quotient will be the time in years; if there be a remainder, multi- ply it by 12, and divide by the rate per cent. 1. For months, thus, 6)100 4 12 16yrs. 6)48(8 months. 2. At 5 per cent., in what time will any sum of money double itself at interest? Ans. 20 years. \ t 4 per cent, in what time will any sum of money double itself, at interest? Ans. 25 years. 4. At 6 per cent , in what time will any sum of money treble itself at interest? Ans. 50 years. I ^ s INSURANCE. 5. In what time will the interest of $240, at 6 per ct., •be treble the principal? Ans 50 years. 6. A certain property valued at $1500 rents for $132 annually, required the rate of interest? Ans. 8f per ct> INSURANC i: :. MARINE INSURANCE. The subjects of Marine insurance are, ships, merchan- dise, freight, &.c. The following examples will clearly illustrate the prin- ciples of Marine Insurance; Real Estate, or Property In- surance. 1. Suppose Ezekiel Doreey, of Baltimore, shipped on board the brig Nimble, Farrell, master; and consigned to David Dunham, Commission Merchant, Liverpool, to •sell for his account: 120 bales cotton, cost $6735.00 1000 barrels Flour at $5.00, 5000.00 Shipping expenses paid, 265.00 Amount of shipment and expenses, $12000.00 Before the brig sails, Mr. Doreey is anxious to have his property insured. Now admitting the rate of insurance to be 1$ per cent, premium, and the -cost of the policy $1.25 cts. How is the amount to get insured obtained, so -as to cover all expenses accurately? Rule. — As 100 less the rate of premium is to 100, so is the sum of cost, charges and policy to the amount re- quired, to get insured Cost and charges .$12000.00 Policy 1.25 100— 1* = 98$ : 100 : : 12001.25 .-$12184.01 * The amount to get insured, to cover all expenses; hence, Mr, Doreey would have to pay to the insurance compa- aiy$184 0L t RANCE. 189 REAL ESTATE. Mv prop iiv m Baltimore is worth $30,000, for what amount must I get it insured, so as to cover cost and charges, insurance being 1 per cent, premium. Poli- cy $1.25. Ans. $30,304.30 — of which $304.30 is to be paid to the insurance company. MERCHANDISE. .'J. Effected insurance on my Warehouse and Mer- chandise therein, which cost me $18,000; what sum must I get it insured for, the insurance being 3 per cent., poli- cy, $ 1 .25. Ans. $ 1 8,557.99— of which the Underwriters is557.99. CASE II. When a Commission Merchant ships a cargo, to his cor- respondent, and therefore cost, charges, premium and policy, are nil included. Rile. — Add jfo of the rate of commission to unity, multiply the sum by the rate of insurance, and call the product b, then as 100 less b: 100 more the rate of com- mission : : the sum of the cost, charges, and policy, to the amount sought, which is to be insured. 4. York and M'Allister, Commission Merchants, New Orleans, shipped on board the brig Orleans. Lewis, mas- ter, and consigned to Lewis Laroque, London, for his account, 920 bales of cotton, cost, $64,531.00 Paid shipping expenses, 266.00 Effected insurance of the invoice amount by the American Insurance Company, at 4 per cent., policy $ 1 25. Ans. the amount to get insured to cover all expenses, is $71,024.33. Amount of insurance at 4 percent. $2842.22 mission on the whole, $67,642.22 at ) OOQO , , 5 per cent, is \ ■WO.H Proof $71,024.33 I RANCE. 5. What sum must a policy be taken out for, to cover $2475, when the premium is 10 per ct. Ans. $2" 6. What is the premium on $896, at 12 per cent. Ans. $107.52. 7. A certain company own a cotton factory in Pitts- burgh, valued at $26,250 for what sum must a policy be taken out, to cover cost and charges at 12 £ per cent.? 1 Ans. $30,000. COMMISSION. 1. What is the commission on $850 at 6 per cei, Ans. $51. J. Calculate the commission on $37,702.46 at 5 pel cent.? Ans. $1885.1 J. 3. The sales made by an Auctioneer amount to $209,723; what is his commission at 5 per cent.? Ans. $10,486.15. 4. An Auctioneer's commissions at 5 percent, on sun- dry sales in the city of New York, in 1839, amounted to $13,279.58, required the amount of sales made? Ans. $265,591.60. 5. A merchant having $1728 in the Chesapeake Bank of Baltimore, wishes to withdraw 15 percent.; how much -will remain? Ans. $1468.80. DISCOUNT Is an allowance made on a bill, or any other debt not yet become due for prompt payment. The discount taken from the principal leaves the pre- sent worth, or value of a bill, when discounted. Q. By having the present worth, how is the discount obtained? A. Subtract the present worth from the principal, and the remainder will be the discount. Rule. — To find the present worth at 6 per cent., if ihe time be for years or months; as 100 plus half the .months is to 100, so is the given sum to the present wrojJth. 191 1. What is the present worth of .^1333.20, due 1 year nee? IUustr yr$. m. 1 10 12 2)22 11 100 11 dolls, els. as 111 — 100 133.20 100 lll)1332000($120.0O 111 222 222 What is the discount of $133,30, due 1 year, 10 months hence? Ans. $13.20. Principal, $133.20 Present worth, 120.00 $13.20 discount as required. '}. What is the present worth of a note for $520, due 5 years hence? Ans. $400. 4. What is the discount on the above mentioned note, for $520, due 5 years hence? Ans. $120. What is the present worth of $775.50 due in 4< years, at 5 per cent, per annum? Ans. $64-6.25. What is the discount of $802.50 at 7 per cent, due one year hence? Ans. $52.50. 7. What is the discount on a note of $117.60, due 1 year hence, at 12 per cent? Ans. $12.60. 8. What is the present worth of $1350, due 5 years, 10 months hen. Ans. $1000. 192 BARTER. BUYING AND SELLING STOCK. 1. What is the amount of $156* United States Bank Stock at 114 per cent? Operation, $1564 114 6256 1564 1564 $1782.96 Answer. 2. Sold 15 shares $100 each, of the Marine Bank of Baltimore, at 13 per cent, advance, what is the amount? Ans. $1695. 3. What is the value of 10 shares in the Philadelphia and Trenton Railroad stock, at 85 per cent, original shares being $100? Ans. $850. 4. What must be given for 8 shares in the Baltimore and Ohio Railroad stock at 10 per cent advance, the ori- ginal cost being $100 each? Ans. $880. 5. What will $1686 corporation notes of the city of Wilmington, Del. be worth at 91 J pr. ct. Ans. $1542.69. BARTER. When one commodity is traded for another, it is called Barter. 1. How much sugar at 10 cents per lb. must be given in barter for 750 lbs. of raisins at 6 cents per lb.? Ans. 450 lbs. Illustration. — 750 lbs. raisins at 6 cts. is $45. Now the question is, how many lbs. of sugar must we get for $45; divide $45 by 10, and we get 450 lbs.; hence, 450 lbs. sugar at 10 cts. is equal to 750 raisins at 6 cts. 450 X 10 - 750 x 6 = $45. 2. Aaron Abel, bought of Ben. Bailey, 6 hogsheads of rum, containing 410 gallons, at $1.17 per gallon, and BARTER. 193 353 pounds of coffee at 21 cents per pound, in part of which he pays $21 in cash, and the balance in boards at $4 per thousand, how many feet of boards did the balance require? Ans. $127957 3. Richard Rich has 240 bushels of rye, which cost him 90 cents per bushel, this he trades with Peter Parley at 95 cents for wheat, that stands Parley 99 cents per bushel, how many bushels of wheat is he to receive in trade, and at what price is it to be rated in order to make the barter equal ? Ans. 218^ V * V bush, at $1.04 J cts. per bush. 4. A. has sugar which he barters with B. for 4 cents per lb. more than it cost him against tea, which cost B. 40 cents per lb., but which he puts in barter at 50 cents, what did A's sugar cost him per lb. Ans. 16 cts. 5. A merchant delivered 3 hhds. of wine at $1.10 per gallon for 126 yards of cloth, what was the cloth per yard? Ans. $1.65. 6. A farmer traded 20 bbls. flour at $5.60 per bbl., for salt at $3.50 per barrel, how many barrels of salt must the farmer receive for his flour? Ans. 32 bbls. 7. How much corn at 45 cts. per bushel will pay for 33 yds. cassinet at $1.05 per yd? Ans 77 bushels. 8. D. Greves has 100 cords oak-wood for sale, worth $4 per cord cash, and is offered $4.25, payable on a cre- dit of twelve months, which is the most advantageous, the cash or credit sale? Ans. the credit by 94 cents, al- lowing discount at 6 per cent, per annum. 9. A. gave B. 2 hhds. of brandy at 75 cents per gal- lon, for 56 yards of cloth, what was the cloth per yardt Ans. $1.68|. 10. What quantity of tea at $1.30 per lb, must be B for 2,500 lbs. of rice at 4 J cents per pound? Ans. 86 lbs. 8 ozs. 1 1 . A grocer had sugar at 8 cents per lb. for some of 17 194 BARTER. which, B. gave 750 lbs. of tea at $1.08 per pound, how 7 many lbs. of sugar must B. receive for his tea? Ans. 90cwt. 1 qr. 17 lbs. 12. John Doe, bought of Richard Roe, 104 tons of iron at $10 per ton, and is to pay as follows, viz: in cash $600, 240 lbs. sole leather at 33 J cts. per lb., and 10 loads of coal, each containing 12 bushels at 37| pr. bush. 90 gallons of brandy at the rate of $75 per hhd., and the balance in coffee at 12^ cents per lb., how much coffee is Richard Roe to receive? Ans, 1662$. 13. Two persons A. and B. barter, A. has 17 cwt. of hams at 12f cents per lb., B. has 1400 lbs. cheese at $10 per cwt; which of them must receive money and how much? Ans. A. $113. 14. A man exchanged 40 bushels of salt at 87£ cents per bushel for 200 bushels of oats at 18J cts. per bushel, how much was the balance in his favor? Ans. $2.50. 15. A farmer sold a grocer 20 bushels of rye at 62 £ cts., 200 lbs. cheese at 10 cts. per lb.; in payment he re- ceived 20 gals, of molasses at 25 cts. and the balance in cash, how much money did he receive? Ans. $27.50. 16. A farmer sold a grocer 15 barrels of apples, each bbl. containing 3 bushels, at 40 cts. per bushel, 30 bush, of corn at 90 cts. per bushel, 500 lbs. of cheese at 8 cts. per lb., 200 lbs. of butter at 15 cts. per lb., 20 bushels of onions at 75 cts. per bushel. In payment he received 4 barrels of Monongahela whiskey, each 31 1 gallons at 33 a cents per gallon, 80 gallons of molasses at 35 cents per gal., 4 bags of Laguayra coffee, each containing 150 lbs. at 10 cts. per lb., what is the balance? Ans. 0. 17. A. has B's note for $535, payable in 14 months, and to redeem it for prompt payment, gives him 160 bushels of corn at 45 cts. per bushel, 122 do. wheat at $1 per bushel, 80 of rye at 90 cts. (market price,) and the balance in brick at $10 per thousand, how many thousand must A. deliver? Ans. 23 thousand. LOSS AND GAIN. 195 LOSS AND GAIN, Is one of the imaginary accounts in Double Entry Book Keeping, and so called in order to supply the want of real or personal titles in recording gains and losses, which could not with propriety be placed to real or per- sonal accounts. It is by this Rule, that merchants dis- cover the gain or loss in business, and the rate per cent, in buying or selling goods. Q. What does the words buy or bought, sell or sold im- ply in this rule? A. Buy or bought, means the first cost of the goods, r sold, the sales made. Q. What does the first cost or invoice refer to? A. The debtor side of a merchandise account Q. What does the items sold refer to? A. The credit side of a merchandise account. Q. How is a gain ascertained? A. By taking the difference between the first cost and ales. Q. How is a loss ascertained^ A. When the amount sold is less than the first cost, rake the difference, that difference will shew the loss. First cost and a gain given to find the rate per cent. Rule. — Take the difference between the first cost and the sales. Then say as the first cost is to the difference, so is 100 to the rate per cent. I 1. Bought coffee at 10 cts. per lb. and sold it for IS cts., how much is the gain per ct? Ans. 20 pr. ct. . Sold for 12 Bought for 10 M 10 : 2 gain : : 100 = 20 pr. ct. Ans. 196 LOSS AND GAIIS. CASE II. Given the first cost and a loss to find the rate per cent. Rule. — As the first cost is to the loss, so is 100 to the rate per cent. Ex. 2. Bought sugar at 8 cts. per lb. (which was dam- aged,) and had to dispose of it at 5 cts. per lb., how much is the loss per cent? Ans. 37| pr. ct. Bought for 8 cents. Sold for 5 " As 8 : 3 cts. loss : : 100 = 371 pr. ct. Ans. 2 25 CASE III. To gain a certain per centage. Rule. — As 10 per cent is T ' v of a hundred, add T * v to the first cost. Thus, T ^ of $4.00 is 40 cts. which added, make $4.40, the selling price required; or, agreeably to the common method, say as 100 : 110 : : $4.00 = $4.40 as before. The first method is more concise, and there- fore, the best. 1. Admitting I purchase cloth at $4 per yard, and wish to gain 10 per cent, how much is the selling price? 2. Bought calico at 20 cts. per yard, and want to gain 25 per cent, how much must be the selling price per yard? Ans. 25 cts. CASE IV. Setting price ^gain andrateper cent given to find the first cost. Rule. — Add the gain per cent, to 100, then say as 100 with the given per centage added, is to 100, so is the re- tail price to the first cost. 1. Suppose the retail price to be $5 per yard for cloth, and the gain 25 per cent, how much is the first cost? Ans. $400 per yard. LOSS AND GAIN. CAS I When a certain loss per cent, has to be sustained on the first cost to find the retail pr le. — Multiply the first cost by the given per cen- tage, and divide by 100 for the amount of loss sustained, unount subtracted from the first cost will give the result; or, as 100 : 84- : : 8*0 : $705.60 as required. Suppose I purchase a lot of goods for $840, at auction, and owing to their being damaged, have to sustain a loss of 16 per cent below the first cost, how much will they be sold for? Ans. $705.60. CASE I I To find the profit per pound, per yard, Sfc. when the first cost and sales of a certain number of yards, pounds, cwts. Sfc, an Rule. — Find the whole gain, and divide it by the number of yards, pounds, or cwts., &c. Ex. 5; Bought a chest of tea, containing 340 lbs. for », and sold it for $408, what was the profit on each pound? Ans. 35 cts. per lb. CASE TIL. I there is a loss per cent, sustained by the sale und the first cost is required. Rule. — Subtract the rate per cent, from 100, then say 1 >0 less the rate per cent, is to 100, so is the amount ic sale to the first cost. Ex. 6. Sold a lot of dry goods for $540, and had to i a loss of 10 per cent., what did they cost me? Ans. $600v CASE VIII. When you purchase goods and wish to gain a certain sum on the whole purchase, so as to be able to sell them at a certain per centage. • . Rule. — Add the sum you wish to gain to the amount of the purchase, then divide that amount by the number of jards, pounds, and cwts. for the answer required. 17* 198 LOSS AND GAIN. Ex. 7. A merchant tailor bought 100 yards of cloth for $256, how much must he sell it per yard to gain $44 on the whole? Ans. $3 per yard. 1. Bought knives at 20 cents each, and sold them for 25 cents, how much is gained per cent.? Ans. 25 pr. ct. 2. Suppose I buy cloth for $5 per yard, and sell it for $6, how much is the gain per cent.? Ans. 20 pr. ct. 3. A store-keeper sold 100 yards of silk at $1.50 per yard, which cost him $1.25 per yard, how much did he gain by the sale? Ans. $20 on the whole. 4. George Brown, merchant, Baltimore, purchased the following bill of goods, viz: 10 pieces domestic muslin, each 29 yards at 10 cents per yard, and 16 pieces calico, each 27 J yards at 20 cents per yard, how much is the re- tail price per yard on the muslin and calico, and the whole gain allowing a gain of 25 per cent, for the sale of the goods:* A. The retail price of the muslin is 12 J cts. per yard, and the calico 25 cts. per yard — whole gain_ $29.25. 5. Sold a yard of cloth for $1.55, by which was gain- ed at the rate of 15 per cent.) what would have been the gain per cent, if it was sold for $1.72? Ans $27.61 -f 6. A. sold cloth at 84 cents per yard, and gained 10 per cent., should it be sold for $1.20, what would be the •.rain' Ans. $57 j. 7. Richard Rich purchased of D, Draper, 18 pieces of broad cloth at $8 per piece, of which he sold 10 at $9.60 per piece, 5 pieces at $9 per piece, and the balance he wishes to dispose of, on terms to gain 12 per cent, on the whole, at what rate must the remainder be sold at per yard? Ans. $6.76. 8. Sold a lot of dry goods for $5.50, and gained 10 per cent., how much wis the first cost? Ans. $5.00. 9. Bought at auction a hhd. of molasses for $50, and having found, that there was a leakage of 5 gallons, I sold LOSS AND GAIN. 199 it imi , at 10 per cent, loss, how much did 1 re- Ans. $45. 10. Bought rum at $1.25 per gallon, and by accident, so much leaked out, that I am willing to lose 20 per cent., at what price must I sell it per gallon? Ans. $1.00. 11. Bought for cash $1740 worth of hardware, and sold the lot for $2000 on a credit of 3 months, what is the gain per cent, per annum? Ans. 59}|. 12. Bought goods to the amount of $1400 cash, and sold them at 4 months credit for $1600, what is the gain per cent, per annum? Ans. 42$c. 1 S. Bought 3000 gallons of molasses at a credit of 2 months, for 30 cents per gallon, and sold it again at 35 cents per gallon, on a credit of 3 months, required the gain per cent? Ans. 64§£c. EXERCISE QUESTIONS. 1. I purchased 100 boxes of prunes at $2.10 each, and by selling them at $3.50 per cwt. the gain is 25 per cent , the weight of each box is required? Ans. 84 lbs. J. If I sell linen for 75 cents per yard, and gain 12| per cent., what would the prime cost of 30 pieces, each 33} yards, stand me in the whole] Ans. $670. 3. Bought a quantity of leather at 22 cents per lb., how should I sell it to gain 10 per cent profit? Ans. 24c. 2m. 4. Admitting that a yard of muslin was sold for 32 cents, which cost 37} cents, what is the loss per cent? Ans. 141. 5. Sold 500 penknives at 17 cents a piece, and by so doing lost 9 per cent, what is the loss on the whole? Ans. $8.40 -f 6 Bought a hhd. of wine for a certain sum, but 15 gals, having leaked out, I sell the remainder for $2.21. 6} per gal., and thereby lose 5 per cent, on the cost, what was the cost 3 Ans. $112. 200 COMPANY ACCOUNTS. COMPANY ACCOUNTS. Partnership or Company transactions, (in an Arithme- tical sense) is a Rule by. which merchants in partnership adjust their accounts in proportion to stock and time. The gain or loss, witl^jthe several sums at hazard given, to find each partners share. Rule. — Multiply the whole gain by each man's frac- tion part of the stock— or, agreeably to the Old Method; say, as the whole stock is to the whole gain, so is each man's share of the stock, to each man's share of the gain. 1. Three men, A, B and C, entered into partnership for 2 years, with a capital of $6000, A put in $2500, B $1500 and C $2000, they gain $1080; required, each man's share of the gain? Solution. — As the whole stock in trade is $6000, A's share would be if ft — At B'* i*H = *> and Cs Jft| = f Again, T ' T of $1080 — $450, A's share, and i of $1080 = $270, B's share, and C's J of $1080 = $360. Proof. — A's share of gain is $450.00 B's do. do. 270.00 C's do- do. 360.00 Whole gain $1080.00 2. A and B enter into partnership; A has in goods at cash price $3400 worth, and cash $1300: B puts in $1200, and agrees to pay for A, a debt of $1100, for which A gives B a title to that amount of his goods. Now suppose A, agrees to take B's note for what B's funds want, of being equal to his own, (say the note bears legal interest, and is not reckoned in the partnership.) COMPANY ACCOUNTS. 201 what amount should the note be drawn for, to make them equal? Solution.— Amount of A's goods $344)0.00 " Cish 1 300.00 4700.00 A sells to B, goods for 1 100.00 B's cash $1200.00 Goods 1100.00 3600.00 2300.00 $1300.00 difference. A's stock in trade then would be $3600—650 » $2950. A and B's equivalent composed of (Goods $1100.00 J Cash 1200.00 f Note 650 00 $2950.00 proof. CASE II. Apportioning the effects of a Bankrupt. In apportioning the effects of a Bankrupt amongst his creditors, it is more convenient to find the proportion for one dollar, &.c. Which will be a constant multiplier for each debt. 1. A Bankrupt owes $5000, his effects sold at auction, amounting to $4000, what will his creditors receive on a dollar? Ans. 80 cents. J. A merchant having sustained many losses is obliged to become a bankrupt, his effects are valued at $1728, with which he can pay only 15 cents on the dollar, what did he owe? Ans. $11520. 8, A, B and C, freighted a ship with 106 tuns of wine, of which A had 48* tuns, B 36, and C 24, but by reason of stormy weather were obliged to cast 45 tuns overboard; how much must each man sustain of the loss? Ans. A ^0 tuns, B 15 and C 10. 202 COMPANY ACCOUNTS. 4. Three merchants trading together lost goods to the value of $1860. A's stock was $2280, B's $11520 and C's $4-800; what share of the loss must each man sustain? Ans. A $288, B $1152, and C $480. 5. A ship valued at $25200 was lost at sea, of which £ belonged to A, i to B, and the remainder to C; what is the loss on $1.00, and how much will each man sustain, supposing the owners effected an insurance of $18000? A's $2400, B's $3600, and C's 1200. The pro-rata share on a dollar is f . case III. When stocks have been put in trade for different periods of time, and settled with regard, both to stock and time. Rule. — Multiply each man's stock and time, and then as the aggregate of products is to the whole gain, so is each man's stock to each man's share of the gain. 1. A, B, and C, join in company: A's stock is $100, for 12 months, B's 120 yards of cloth, for 8 months, and C's 240 bushels of wheat, lor 7 months; they gained $1612, of which A had $400, B $512, and C$700; what was the value of B's cloth per yard, and C's wheat per bushel. Ans. B's cloth $1.60 pr yd, and C's wheat $1.25 prbu>h. 2. A, B, and C, enter into partnership with a capital of $1100, of which A put in $250, B put in $300, and C $550; they lost by trading, 5 per cent, on their capital, what was each man's shaie? Ans. A's loss $12 50, B's $15, and C's loss $27.50. In company accounts, when the times and payments are equal, the shares of gain or loss are evidently in propor- tion to their respective stock — and when the stocks are equal, the shares are in proportion to the times of pay- ment. But when stocks and times are unequal, the shares are in proportion to the products of stock and time. This may be clearly demonstrated thus: Suppose $100 in trade 12 months, gain $20; $50 in trade in 6 months, will gain $5, and both together $25; EQUATION OF PAYMENTS. for, as 100 X 12 : 50 x 6 : : 20 : 5 and 20 + 5; again, by composition 100 X 12 -f- 50 X 6 : 100 X 12 JO; gain of $100 in 12 months, and 100 x 12 -f 50 x 6 : 50 X 6 : : 25 : 5,gain of $50 in 6 months, from which the truth of the rule is evident. A, B and C having traded together, gain $126.80 — what is each man's share, allowing that A put in $50 for 4 months, B $100 for 6 months, and C $150 for 8 months? Am. A $12.68, B $38.04, and C $76.08. EQUATION OF PAYMENTS Is a rule, for finding when any number of notes or bonds due at different times may be all paid at once, without loss to debtor or creditor. Rule. — Multiply each payment by its time, divide the ram of the products, thence arising by the sum of all the payments, and the quotient will be the equated time required. 1. A. owes B. a bond for $100, due 2 months hence, and one for $500 due 18 months hence, what would be the equated time for paying them at once? months. Operation. $100 X 2 = 200 500 x 18 - 9000 mo. 6)00 6)9200(= 15 1 Ans. 2. C. owes D. $550, of which $100 is to be paid at three months, $200 at 5 months, and $250, in 8 months, hut have agreed to make one payment of the whole, at what time must it be paid? Ans. 6 mos. "J. A man has owing to him $500 to be paid as fol- lows, viz: $250 at 6 months, and $250 at 8 months, but it is agreed that the whole should be paid at one time, when must it be paid? Ans. 7 mos. 4. A. owes B. 5 bonds, for $945 each, payable at 3, 9, 11, 19, and 29 months, what time might they be all paid at once? Ans. 14$ mos. '2(H GENERAL AVERAGES. AVERAGE TIME OF SALES. CASE I. 1. Sold merchandise at sundry times, and on different terms of credit, as per statement annexed. 1840 January 1st, $1500 at 3 months, due 1st April. February 10th, 250 at 2 " " 10th " March 19th, 643 at 4 " " 19th July. Sept'ber 1st. 1400 at 6 " " 1st March. Required the average- time of payment. Ans. 21st August. Statement of the preceding question. days. Due 1st April $1500 x " 10th " 250 x 9 « 19th July 643 x 109 " 1st March 1400 x 334 3793 3793)539937(142 142 days from the 1st day of April which will make the average time fall on the 21st day of August Theorem and General Rule, to find the average time thai several bills of different dates, or different terms of credit, or both, become due. W « _1 D C B In the above diagram let p, q, r, s, t, &c, be the several payments to be made, and B, C, D, E, F, &c, *he different periods of time at which those payments are to be made, and O, the average point of time, then it is manifest (on the principal of Simple Interest, that p X BO + q X CO + r x DO = t x FO -f s x EO, and putting BO = x, we have p x + q (x — BC) -f r ( x — BD) =s (BE— x) -f t (BF — x) by transposition \ERAL AVERA(. 205 px+qx + rx + si + txoqxBC + rxBD -f 8XBE+txBF(p-fq-f-r+s-ft)x=x — g x BC -f r x BD 4- s x BE + t x BF. p + q + r + s+t. Hence the following general rule: multiply the several payments to be made by the respective times from the first payment, add them together and divide that sum by the whole amount of the bills for the time sought, which is to be counted from the time on which the first pay- ment falls due. SOLUTION OF QUESTION I. CASE I. 334 1st March x 19th July. 10th April. 1st Ap. I * 1 1 1 Let x = aver- age time from 1st April, then 1500 x + 250(x — 9) + 643 (x — 109) = 1400 (334 — x) then 1500 x -f 250 x — 2250 + 643 x — 70087 — 467600 — 1400 x. Again, 1500 x + 250 x+ 643 x + 1400 x =467600 -f 70087 . , 2250 + 70087 + 467600 , . , -f 2250 and x = , — ! ^ = 142 days after the 1st day of April, which agrees with the 21st day of August, 1840, as before. SOLUTION OF CASE I. BY ANOTHER PRACTICAL METHOD. T > T of the first bill is $150 x = 000 " # 2d " 25 x 9 = 225 " 3d « 64 x 109 — 6976 " 4th " 140 X 334 — 46760 379 379)53961(142 Note. — Agreeably to mercantile usage, a fraction less than one-half in dollars or days is omitted in the equa- tion of payments, and when more than a half, it is consi- dered as a unit. 18 GENERAL AVERAGES. case ir. 1. Sold merchandise at different times, and on rtriom terms of credit. 1839 September 6th $100 for 1 month, due Oct. 6th. * 14th 125 " 1 " u " 1 i tli. October 10th 175 « 2 " a Dec. loth. November 14th 340 " 3 " a Feb. 1 1th. January 1 1th 456 " 5 « u June 11th. FIRST METHOD. Solution. days. October 6th $100 x " 14th 125 x 8 Dec'ber 10th 175 x Feb'ry 14th 340 x 131 June 14th 456 x 251 1196 1196)171371(=M1 + 144 days from the 6th October will come to February 27, 1840, at which time a note for $1 196 would be due. SECOND METHOD. October 6th $100 X 231 « 1 hi. L25 x 243 Dec'ber 10th 175 x 186 Feb'ry 14th 340 x 120 June 14th 456 x 000 1196)128825(=10S -f $1196 Due by average 108 days earlier than thqgl4th day of June 1840, which will be the 27th day of Feb. as above. THIRD METHOD. 1839 September 6th $100 X 30 " 14th 125 x 38 October 10th 175 x 95 November 14th 340 X 161 January 14th 456 X 281" g 1196 1196)207251(174 -f 1 74 days from the 6th September, which will make the average time fall on the 27th day of Feb. 1840, as before. \l. AVERAGES. 207 i K AGE CALCULATIONS BY INTEREST. days. mbcr (ith $100 X 30 at interest 14th 125 X 38 " October I Oth 17") X 95 « November 14th 340 x 161 " January Hth 45b' x 281 « Interest, $34.57 $11.96 : 60 : : $34.57 — 174 days, nearly being i in- required for $1196 to gain $34.57, the whole time by equation from the 6th Sept. to the 27th February. CASE III. To average personal accounts when goods are recein hurt i r. I. le. — Find the average time for the debtor and cre- ditor sides of the accounts, and multiply the least sum by the difference of time, divide the product by the dif- ference of the accounts, and the quotient will be the number of days to be carried forward. Illustration. — Sold Barton & Co. Philadelphia, mer- chandise at different dates and credits for $600 due by average 6th September. Received from Barton & Co. at different times, goods amounting to $1000, due by average on the 6th January, 1841, required the time of payment for the balance due the firm? From $1000 Least sum $600. Take 600 400 Difference of time from 6th September to 6th January. 600 x 91 400)546(00 136 J days. 208 GENERAL AVERAGES. Due by average 136 J days from the 6th day of Janua- ry, which falls due on the 23d day of May. Dr. Barton & Co. Phila. 1840. Sept. u'th to Mdse. As per average, $600.00 iMay 23d, balance, 400.00 When due, $1000.00 Barton & Co. Cr. 1841, **£:!*«»•«> 1 000.00 From the 6th September to the 23d May (average time) is 227* days, and 600 x 227* — 1000 x 136* = «»y«» mt $22.75 equated interest for Dr. and Cr. CASE V. ANOTHER METHOD. I. A. B. purchased merchandise on credit at different times, and is desirous to pay the whole amount at a cer- tain period, required the time by equation? Sales made as per bill rendered. dolls. days. product. 1840 May 4th $52.00 X 4 " 8th 13.87 66 x 29 • June 6th 104.20 170 X 4 " 10th 84.16 254 x 34 # July 14th 125.04 000 x 379)11438(30 Thirty days counted back from July 14th will give the equated time on the 14th of June. EXCHANGE. 209 EXCHANGE Q, What is Exchange? A. It is the paying or receiving money in one country, ;s equivalent in the money of another country by means of Bills of Exchange. Q. What is meant by a Bill of Exchange? A. A Bill of Exchange is a written order for the pay- ment of money at an appointed time. Q. Who is the drawer? A. The "maker and seller" of the bilL Q. Who is the acceptor? A. The person to whom the bill is addressed, is called the acceptor; when he engages, to pay the bill. Q. What is meant by endorsing a Bill? A. When the holder of a bill, disposes of it, he writes his name on the back, which is called endorsing. Q. Are there other persons occasionally concerned in a Bill of Exchange? A. Yes, the buyer or remitter, the seller or negociator, and the holder or possessor. Q. Who should be the first endorser to a Bill? A. The payee. Q. What is a draft called, when a Merchant in the U. S. draws on his Banker in London? A. "Bill on London," and vice-versa. Drawee. — The person to whom a bill is addressed. Payee. — The person to whom a bill is ordered to be paid. Indorsee. — The person to whom a bill is made payable. These are technical terms in too. \^' 210 CCH o ANGE. ^ O O O <N O 00 h » h o W CI •* CO T* T-* H O O •/.• -*-» i i § 3 i .« e . -a • 3 « ^ • | i Si 4» vereij orD • o O O • • £3 It * It 1 i Guinea < Congres ole, nea, wn, pain, Sw a * ^ a 1 £ cg 3 nglish ict of hPist hGuii h Cro rof S] -a .2 • 2pu a°-S j *- o o o 5 w o a a a -5 a 2 a~ £ £ Co < fcfefeQ-^cZ) B 30000O0O»w •J 9 oooooo p © p © o o O O O 00 G> O O C> 3 / — -i — ;o co *-5 -* r- — r-i •§ • • the United Si i i i i i i i i a • ' jt •= a • i : v a es, valu e sixtee i i i i i 5 o o d o S . . a -O T3 T3 T3 ° o o -a t3 uble Jo gle If arter ;hth stoon,o •more, If arter ubloon EXCHANGE. 211 Bi-NifjoOOOO O© © *-« »fi <N © © dddoHwco ^ <N f- »rt O ex as r» in © o >o £ f# « c «? — a) 3^ o a - • : o o o o *-i ■ee- « •-» 3 =: ~ ^ « fcOOJ .© OC C* 00 o o EQO ~ OS 00 CO CO a© co o co oi <n oOOH«hO) d d d d d d c* o -H O co d #x P «J O » O O 'o £?■£•£ fc ca -•-» 2 « > « c o -C .- a> i-, a> a o 'o (NX-* i> oo co d oji oo o o o odd ■fie- C* CX o p lO Oi t> *T ^ »ri co t^ ,-< (N © © d d *i c4 * e~ OPm > P-. > aoOao £ 2 M ° uuSQ *S x .* >* CO CO 99 "' - X sd d d d d d o d d d ih rn. co d X 8 "3 * .2 * O 3 M M i-H co co o »o co OOHOjWW © © © © © © 3 sps.s.s 1 d co o oo t4oo o +r .H 3 cGO O* C* O f- © «* O = OC500'--f<J>t^ S'OOOOOiO^OOCO uOOH-OOrjlHft OO oB'-i'-t^O© ^ ^o of £ • u i © > - FRANC r 12 deniers, Tumois or 2 r Crown 6 Li e 10 livres, d'or - i < I to" B i i = enie ol, ivre cuo istol ouis a <u £ > Qwjwa.j^fH r* ~* O O CO O o* CO 00 o oo p ^4 r4 GO t^ £• O O *-« CO Oi p p d c4 oo 4^ i i O J0 y tfl o *3 •- 3 "o t»12 MERCANTILE FORMS. ^\ i 4 M 4 3 II a o »— i 1 o 2 3 CO -2* "a 9 > c 3 PS HE FORMS. 213 00 r 15 i IC I C 6 k 6 Si o *> •a i *> if bo I s I k S •I si , ■a -5 © a 1 *© a I 8 £ 21 1 MERCANTILE FORMS. & 9 S 9 9 § ^5 1 8 ft! &3 ^ I 6 s >3 ■*• ^3 ~» <N s I s •I i CO « o o :* & . 5S ^ 8 bearc Ada ** ** u o =~ •^ 1 *» 1 cq tt •*5 -: i* o *"» •"» B»2 s *-rf •S s 0fd %i cc *) • — - ■— S ,3 £^ a, c *" £ a s 52 ^~ *s^ CO <*> ^ s> ■*«* • ** mon rece H S & L, ^ c*T k 3 o ^ 35 IF.E FORJfS. © >> *) •o 5 4 g I 1 r . o • * s * a J © -< 5 .2 9 s I 16 i 4 MERCANTILE FORMS. $ >3 ^ I e? v> L*\ h V c ©o ^ i *s MKRCAN'Tii m K E M i 5 3 i 5s ^ -I N ^ ^ SI ^ si il 1 * 1 &S 19 218 MERCANTILE I'oRMs. a 9 & 9 a -I 1 B -< ;• OS 1 Q « -*: 5 s r* §>■?•§ K)^ -S ^ ^ s - <S s: ^ 1 95 ~J d a* ^ ftri k S S «*, ««». »■»• >- -s* ^ £^ If i i ft; | . * |j MERCANTILE FORMS. 219 *c O* ^j -— •— S o* ?• ^ t I ~ * s =5 ^ l r, i •$ « j < I I 2 ^^"^bc^o^^ a is ■* * *• *-s it rl tl ;•] £3 I rfS * J - <S ^ n r ^ § ^ « • I S^ 5 <"« If £ =S (§ « "*• P, ~ § S « & * .8 "j •« * * * £ « -s Moil ^«^ yf »% lit * 8 ^ h I i * * . *"* -■ « ^ .^ o 6 5 1 4 *s 2 * ■* 4; $ iSj no MERCANTILE FORMS. ¥ § i 2 M n j M R i 1 5: 1 it 41 s CI 221 When Sterling money is required. 1. Suppose A. Allen of Baltimore, has occasion to on B. Burton of London, for $5462.50 when the iange is at 9\ per -cent, (in favor of England) it is re- < d to find how much sterling money the bill must be drawn for? « Ans. $5000. //A.— As 109.25 : 100 : : $5462.50 : $5000. To reduce $5000 American money to sterling, multi- ply the dollars by 9, and divide by 40 (i, e.) 5000x0 = £ I 1 29j the amount required. 40 <^. Why do we multiply the dollars by 9 and divide by 40? A. Because, when the dollar is at par, 4s. 6d. make a dollar, and as 9 sixpences make a dollar, and 40 a pound sterling) therefore we multiply the dollars by 9 and di- vide by 40. CASE II. To change Pounds Sterling to Dollars. Ri i.e.- -Multiply the pounds by 40, and divide by 9, and we get the answer in dollars — then as 100 London : 109.25 Philad. : : 5000 Lon. : $5462.50, the amount t in federal money. 1. A merchant in Philadelphia has occasion to draw OQ Paris for the above sum of $5462.50, when ihe ex- change with France is at 5 francs, 22| centimes per dol- lar; — required, how much French money the bill must be drum for? Ans. 28541. 56* francs. CASE III. iange 5 Franc Pieces (French money) to Dolt 1. — Say as the value of the 5 franc, agreeable to the rate of Exchange in French money is to $1.00 Amer- ican, so is the amount of the bill in French money to its value in American currency. 19* 222 EXCHANGE. 3. What is the value of £948 16s. 8d. sterling in Amer- ican currency? Ex. As the value of a dollar in Great Britain, may be either below or above par, that is, more or less than 4s. 6d. the proper method is, to bring the pounds, shillings and pence to pence, and divide by the pence in a dollar, for the amount required, thus suppose the dollar to be at par, £ s. d. then 948.16.8 90 *. d. 4 6 18976 12 54 )227720($4217.04 nearly. Answer. 216 117 108 92 54 380 378 200 Note. — Two ciphers are added 216 in order to find cents. 4. How many milreas of Portugal are equal to $12500 American currency? Ans. 10,000 milreas. 5. In 10,000 guilders of Holland, how many dollars? Ans. $3880 U. S. currency. 6. In 10,500 crowns of Denmark, how many dollars? Ans, $7000. 7. In 400 livres turaois (or 20 sols) of France, how many dollars? Ans. $7408. 8. In $8000, how many rials of Spain? Ans. 80,000 rials. BXCHANGEJ 223 .'». In >ii,000 U. S. currency, how many testoons of Portugal? Ans. 48,000 testoons. 10, How many pistoles of Spain are in $36,000? Ans. 10,000 pistoles. 11. A of Amsterdam is indebted to B of Baltimore 3000 guilders, what is the amount, at 38 cts 8 m. per guilder? Ans. $11 64. 1J. How many dollais will purchase a draft of 2500 Italian pistoles? Ans. $8000. BANOE TABLE FOR BUYING AND SELLING BILLS OF EXCHANGE, ABOVE AND BELOW PAR. 1 * W 1 1 ^ it * it 1 ° 4 |<H j IN I h - ^ m * *2 1 1.0025 4 1.04 8i 1.0825 h 1.005 H 1.045 8* 1.085 1 1.0075 5 1.05 81 1.0875 1. 1.01 5* 9 1.09 n 1.0125 51 1.0575 n 1.0925 H 1.015 6 1.06 H 1.095 M 1.0175 6* 1.065 9| 1.0975 1 1.62 «l 1.0675 10 1.10 2i 1.02H5 7 1.07 101 1.1025 2* 1.026 n 1.0725 10$ 1.105 2| 1.0275 n 1.075 10J 1.1075 3. 1.03 1.0775 11 1.11 3* 1.035 8 1.08 111 1.1125 This Table is convenient for finding the amount of bills at any rate per cent from i to 1 1 4 per cent., above or below par. CASE IV. To find the amount of a Bill of Exchange when the rait per cent, is above par. 1. What is the amount of a bill of exchange for $750 At 4 per cent, above par* Ans. $780. 224 EXCHANGE. Illustration. Opposite to 4 per cent, in the annexed table, the tabular number is 1.04 X 750 = $780. J. What is the value of a bill of exchange for $1175 at 9f per cent, above par? Ans. $1289..%.',. 3. What is the value of a bill of exchange for $5333 at 7 per cent, above par? Ans. $5706.31. CASE V. To find the amount of a Bill of Exchange when the rate per cent, is below par. 1. What is the value of a bill of exchange drawn for $9630 at 7 per cent, below par? Ans. $9000. Rule. — Divide the given sum, for which the bill i« D for, by the tabular number, opposite the rate per cent., thus 9630 — 9000 1.07 8. What is the value of a bill drawn for $530 at 6 per cent, below par? Ans. $500. 3. James Nevins, Stock and Exchange Broker, (Ex- change Place, Philada.) sold the following drafts. No. 1 , $450 at 6 per cent, above par, 2, 900 " « 3, 750 " « Required the amount? Ans. $22*J6. a m u it I •c 3 «■ - «s 3 ! o »a © m © « © »« o »o © »t o o c o ©*©©'©' © © © ©' ©* ©' © ©" ©' ©' ©' ©' i- x ?. c - « « t n x nod a © © © © w =:= 7T«T^Xi — X *5 © 00 © — ?-i re -r x. ? -: r* — O © © © © © c — — oi Ci Ci r: :: o ui © © © © © © © © © © c * © © © © — i-« © © c" © ©'©*©'©" " © ©' © ©' © uE. 225 Emory and Company, Exchange Brokers, Baltimore, sold the following drafts. No, 1, $1500 at 3 per cent, below par 2000 "2 per cent. " " 1875 "2* «* " " Ans. $5246.36 -f CASE VI. Negotiating Bills. 1 . P in Philadelphia owes L of Liverpool £349 1 9s. 3±d. ing, to pay which he buys a bill at 2^ per cent, be- low par, what must he pay in United States currency? Example shewing the use of Table 2d. First put down £349.95 For 3d " 0125 For id " 00208 £349,96458 this reduced to dollars by case 2nd X 40 9)139,98.58320 and 1555.40 $1555.39813=$1555.40nearly 1.025 = $1517.46 4- Answer. 2. A Factor in Philadelphia, owes a merchant in Dub- lin £1500 sterling, to pay which he buys a bill at 4 per cent, above par; how many dollars did that bill cost him? Ans. $6933J. Let these examples be proved, by reducing them to .-ttiling money. 3. A merchant in Bordeaux owes a merchant in Phila- delphia, the nett proceeds of a consignment, amounting; to $750.16, how many francs must he draw for, if ex- change be at 19 cents per franc? Ans. 3948 frs. 21 cts. 1. What must be paid in New York for an invoice of goods, charged at 591 florins 17 stivers, allowing the ex- change at 40 cents per florin, or 2 cents per stiver, and advancing on the invoice 60 per cent.? Ans. $378.78 -f 226 INVOLUTION. CIRCULAR EXCHANGE. CASE VII. 1. London was ordered to remit to Paris 1000 crowns at 32d. sterling per crown, and to draw for the value upon Amsterdam, at 36s. 6d. Flemish per pound sterling; but when the order came up, bills on Paris were at 32*d. sterling per crown, what must be the rate of exchange with Amsterdam to compensate the advance on the remit- tance? Ans. 36s. 2|fd. 2. If the exchange in Hamburgh on London, at 2 usance* be 33, what should it be at sight, reckoning 1 per cent, for the time? Ans. 33s. 4d. 3. Suppose L of London has orders from P of Paris to remit to him at 23 livres 12 sous, (20 sous being equal to one livre) and to draw for the amount on A of Leghorn, at the exchange of 53d per dollar, but L finds the ex- change of London on Paris is 24 livres — At what rate of exchange should he draw on Leghorn to fulfil the order! Ans. 52/ T . 4. A merchant in Baltimore shipped a quantity of flour which, when disposed oi, amounted to 1040 milreas 500 reas, and received in return 17 pipes of wine, how much was the wine per pipe? Ans. $80. INVOLUTON, OR THE METHOD OF RAISING POWERS. 1. The first power of any number is unity multiplied by such number. •J. The second power of a number is found by multi- plying any number by itself. *The term usance is French, and signifies the usage of different countries, in relation to the payment of bills: usances vary from 14 ilays to 1, 2 and 3 months after the date of a bill. SQUARE ROOT. EXAMPLES. Let 2 be raised to all the successive powers, from the the 5th. 2 x 2 sm 4 square of second power. 2 X 2 X 2 =» 8 cube or third power. 2x2x2x2 = 16 biquadrate or 4th power. 2 x 2 X 2 X 2 X 2 — 32, 5th power. 1. A floor is 12 feet square, how many feet of boards does it contain? Ans. 144 ft. 2, In a plantation 300 perches square, how many- perches does it contain? Ans. 90000. TABLE OF POWERS. Roots Squares. Cubes. 4th pow. 5th pow.6th pew. 7th row. 8th pow.9th pr. _ _ _ - -_ _ - I I 9 27 1 32 243 1 64 729 1 m 2187 1 256 6561 512 19681 SQUARE ROOT. 1. To extract the Square Root prepare the number for extraction by pointing it from units place, into periods of two figures each. 2. Find by involution or for convenience sake in the table, a square nearest to the first, subtract and bring down the next period, which place to the right of the re- mainder for a dividend. 3. Double the quotient figure for a divisor, and try how often it is contained in all the figures of the dividend ex- cept the one, on its right. 4. Place this in the quotient for a second figure of the root, as well as to the right of the divisor. Multiply by this quotient figure as in division, the product subtract as before, and to the difference bnng down the third period. SQUARE ROOT. 6. Proceed in like manner, still doubling the quotient figures for a new divisor, and bringing down another period each time for a new dividend, until the whole is completed. EXAMPLE. 1. What is the square root of 10, 34, 26, 50? (3216 9 62)134. 124. 641)1026 641 6426)38556 38556 2. Extract the square root of 151321? Ans. 389 rt. 3. " « « of 2985984? « 1728 rt. 4.. " " « of 230976361 « 4806 rt. CASE II. In extracting the root of whole numbers and decimals, one half the number of decimal figures must be pointed out in the quotient. 1 . What is the square root of 3271.4007? Ans. 57.19 + " " " " of 4.372594? " 2.091 + CASE III. To extract the Square Root of a mixed number. Rule 1. — Reduce the fractional part of the mixed number to its lowest term, and then the mixed number to an improper fraction. 2. Extract the roots of the numerator and denomina- tor for a new numerator and denominator. 3. If the mixed number given be a surd, reduce the fractional part to a decimal, annex it to the whole number, and extract the square root. 1. What is the square root of 421? Ans. 6J. 2. What is the square root of 51 f ±? " 7}. SQUARE ROOT* 229 CASE IV. To i Square Root of a fraction. Rule. — Reduce the fraction to its lowest terms, then extract the square root of the numerator for a new numer- ator, and the denominator for a new denominator. 1. What is the square root of ff ff? Ans. $. % What it the square root of \\%tf " f. CASE V. To extract the Square Root of a Surd. Rule. — Reduce the surd to a decimal, and extract the root thereof. 1. What is the root of |J T ? Ans. 89802 + 1 What is the root of \\\\ « 86602 + CASE VI. To find a mean proportional. Rule. — Multiply the two numbers together, and ex- tract the square root of the product for a mean propor- tional. 1. Suppose A. in a school room sits 4 feet from a hot stove, and B. 9 feet from the same, how much warmer is A. than B? Ans. 6 times. J. Two ships sail from the same port, one goes due North 128 miles, the other due East 72, how far are the ships asunder? Ans. 146.86 miles. CASE VII. The base and perpendicular given to find the hijpothenuse. 1. The top of a castle is 45 yards high, and is surround- ed with a ditch 60 yards broad, what length must a ladder be to reach from the outside of the ditch to the top of the castle? Ans. 75 yds. J. The wall of a fort is 25 feet high, which is sur- rounded by a moat 30 feet in breadth, I want to know the length of a ladder that will reach from the outside of the moat to the top of fne wall? Ans. 39.05 ft 20 SQUARE ROOT. v 1 1 1 . To find the distance that any object may be seen at sea, elevated at any height above the level of the wattr. Rule. — Add to the earth's semi-diaineter in feet, the height of the object; square the sum, next square the number of miles in the earth's semi-diameter, take the difference of those squares, then by (Euclid 47, Lib. 1,) the square root of the difference of those squares will be the distance required in feet. 1. There is a point of the Andes in South America which is 4 miles above the level of the sea, to what dis- tance could a person see from the top of such an elevated point, provided the atmosphere was perfectly clear, and not assisted by refraction. Illustration. — If we put the earth's semi-diameter at 4000 miles, then v/4004 x 4004 — 4000 x 4000 = 178.93 -+• miles, the distance required, which is about twice as far as a person could see elevated from a point one mile above the level of the sea. 2. Suppose a ladder 40 feet long be so planted as to reach a window 33 feet from the ground on one side of the street, and without moving it at the foot, will reach a window on the other side 21 feet high, what is the breadth of the street? Ans. 56, 64 -f feet. Street* CUBE ROOT. 231 1 2 3 4 5 6 7 8 $ 9 1 7 64 (88 216 343 512 729 $ CUBE ROOT. ( Sign #- - ) X Hoots. Cubes. 1. — Point every third figure beginning at the units place, then find the nearest cube to the first point, ami subtract it therefrom; put the root in the quotient, bring down the figure in the next point to the remainder for a dividend. 2. Square the quotient and multiply it by 3, for a divi- sor, find how often the divisor is contained in the divi- dend, rejecting units and tens, and place the number of times in the quotient. 8. Square the last figure placed in the quotient, and place the result to the right hand of the divisor, for a de- fective divisor. 4. Then multiply the last figure placed in the quotient by the other figures, and that product by 30; add the last product to the (defective divisor,) placing units under units and tens under tens, for a complete divisor. Illustration. — Extract the cube root of 4 ^99252847(436 4 (."I. root 4 6 24 30 720 46 3 16 square of 4 352. 5 2 3 33333 138 30 divisor. 191 • 1916847 4140 5556 ^ ^ S oS ~> ~ w Oi O 00 ^H CO 1. What is the cube root of 389017? 2. What is the cube root of 5735339? Ans. 73. * 179. CUBE ROOT. 3. What k the cube root of 32461759? Ans. 319. 4. What is the cube root of 84604519? " 439. 5. What is the cube root of 259694072? " 638. case in. To extract the Oa.be Root of a fraction. Rule. — Reduce the fraction to the lowest denomina- tion. Then extract the Cube Root of the numerator for a new numerator, and also of the denominator for a new* denominator. 1. What is the cube root of ff %1 Ans. f. ■:■ What is the cube root of fflf} Ans. f. 3. What is the cube root of Jf §fc? Ans. |. CASV To extract the Cube Root of a mixed member. Rule. — Reduce the fractional part to its lowest terms, and then the mixed number to an improper fraction, ex- tract the coots of the numerator and denominator for a new numerator and denominator, but if the mixed num- ber given be a surd, reduce the fractional part to a deci- mal, annex it to the whole number, and extract the root therefrom. 1. What is the cube root of 12if? Ans. 2|. 2. What is* the cube root of 31 ^y Ans. 3 j. 3. What is the cube root of 405 T y ? ? Ans. 7f. APPLICATION. 1. If a cubical piece of timber be 47 inches long, 47 inches broad and 47 inches deep, how many cubical inches does it contain? Ans. 103823. 'J. There is a cellar dug, 12 feet in every way, length, breadth and depth, how many solid feet of earth are taken out of it? Ans. 1728. 3. The solid content of a cube is 389017 feet, what is the superficial content of one of its sides? Ans. 5329. CUBE ROOT. 233 Well known principles assumed. Circles are to one another as the squares of their dia- meters. Spheres are to each other as the cubes of their diameters. Cubes and all similar solid bodies as the cubes - ir diameters or homologous sides. Whatever con- es length, breadth and thickness or depth is a solid. EXAMPLES. 1. If a ball 3 inches in diameter weigh 4 lbs., what will be the weight of a ball that is 6 inches in diameter? Ans. 32 lbs. '2. The solid content of a cellar, which is alike in length, breadth and depth, is 100* cubic yards, required the length of its side. Ans. 13.95 feet -f CA8E V. The side of a cube being given to find the side of a cube which shall be double, treble, Sfc. in yuantiiy to the given cube. Rule. — Cube the given number and multiply it by 2, I, \c the cube root of the product, is the side sought. 1. There is a cubical vessel whose side is 12 inches, it is required to find the side of another vessel that is to con- tain 9 timet as much? Ans. 17.306 inches. 2. If a ship of 400 tons burden be SO feet long in the keel; what is the burden of another ship the keel of which is 100 feet long? Ans. 781 tons, 5 cwt. '.\. The dimensions of a ship are, viz: keel 125 feet long, beam 25 feet, depth of hold 15, what dimensions should a ship of similar form have, to carry 3 times the burden? Ans. length of keel 180,28 feet, breadth of beam 3*6.05, depth of hold 21.63. 4. Kind the dimensions of a similar ship that shall con- tain, or carry ju>t hall* the burden of that whose dimen- sions are given. Ans. length of keel 99.21 feet, beam J 9.81, hold 11.09. 291 SINGLE POSITION. 5. Suppose a cannon ball of 4 inches diameter, weighs 18 lb.; what is the diameter of another that weighs 42 lbs? Ans. 5.30 -f inches. 6. Suppose a mortar shell of 8 inches diameter, weighs 50 lb; what is the diameter of a shell that weighs 100 lbs? Ans. 10.08 -f inches. CASE VI. To find mean proportionals between two given numbers. Rule. — Multiply the square of the lesser extreme by the greater, the cube root of the product will be the lesser mean. Again, multiply the square of the greater extreme by the lesser extreme, the cube root of the product will be the greater mean. Example, — Required to find 2 mean proportionals be- tween 4 and" 256. Illustration, 4 x 4 X 256 =* 4096 ^409^== 16. Again, 256 x 256 x 4 =» 262144, the '%^2b , 2144 = 64, hence 16 and 64 are the mean propor- tionals required. SINGLE POSITION. This rule is called Position because by using supposed numbers according to the conditions of the question, the answer is obtained. Rule. — As the sum of the errors is to the given sum, bo is the supposed number to the true one required. Proof. — Add the several parts of the result together, and if it agrees with the given sum it is right. 1. A person, after spending ^ and \ of his money had 60 dollars left; what had he at first? Ans. $144. Suppose he had 24 t 8 Spent 14 then 24—14 = 10 235 DOUBLE POSITION. •! : : 1 (it) •> Us. $144 Proof 1 : i 48 { 36 84 then 144—84 m 60 2. A teacher being asked how many scholars he had, said, if I had as many, half as many, and one quarter as many more, I should have 264, how many had he? Ans. 96 scholars. 3. What number is it whose i, ± and I make 235? Ans. 300. 4. What is the age of a person who says, that if f of the years I have lived be multiplied by 7 and J of them be added to the product, the sum will be 292? Ans. 60 years. 5. A's age is double that of B, and B's is triple that of C. and the sum of all their ages is 140, what is each per- son's age? Ans. A's 84, B's 42, and C's 14 yrs. 6. A certain sum of money is to be divided among 4 persons in such a manner that the first shall have \ of it, the second i, the third J, and the fourth the remainder, which is $28, what is the sum? Ans. $112. 7. What sum at 6 per cent, per annum, will amount to $1032 in 12 years? Ans. $600. DOUBLE POSITION/ Double Position teaches to resolve questions, by mak- ing two suppositions of false numbers. Those questions in which the results are not propor- tional to their positions, belong to this rule. This rule is l the suppositon, that the first error is to the second, as the difference between the true and first supposed number is to the difference between the true and second supposed number. U'li.u this is not tin- case, the exact answer to the ques- tions cannot be found by this rule. DOUBLE POSITION. 1. John asked James how much his horse cost, who answered that if he cost him three times as much as he did, and $15. dollars more, he would stand him in $300, what was the price of the horse? Ans. $95. Suppose 96 Suppose 90 288 270 More 15 More 15 303 285 Should be 300 Should be 300 Too much + 3 Too little— I "> 1 5 IP the errors are in the same ratio as one to five. 96 v 90 96 x 5 = 480 _j- i A 5-!l0xU 90 6 6)570( = $95 Ans. The following rule will be found very concise when the signs of -f or — are alike subtract, when unlike add: Musi rui ion. When both signs are -f- or both signs — , subtract, but when one sign is -f- and the other — add. If the errors are alike, divide the difference of the pro- ducts by the difference of the errors; but, if unlike, di- vide the sum of the products by the sum of the errors. 1. Divide 15 into 2 such parts, so that when the greater is multiplied by 4, and the less by 16, their products will be equal? Ans. 12 and 3. •J. Divide '21 into 2 such parts, so that when the greater is divided by the lesser, and the lesser by the greater, and afterwards the greater quotient multiplied by 5, and the lesser by 125, their products will be equal? Ans. 17£ and 3£. 3. A lady being asked her age thus replied: My age is such, if multiplied by three, Two-sevenths of that product it tripled be; The square root of two-ninths of that is four, Now tell my age, or never see me more. Ans. 28 years. DOUBLE POSITION. 237 4. A laborer was hired 80 days upon this condition, that for every day he was idle, he should pay his employer . and for every day he w^s at work he should re- e 75 cents, at the expiration of the time, he received : now, how many days did he work, and how many was he idle? Ans. he worked 52 days, and was idle 28. Two persons, A. and B. have the same income, A. saves one-fifth of his # every year, but B. by spending $150 per annum more than A. at the end of 8 years finds himself $4-00 in debt, what is their income, and what does each spend per annum? Ans. Their income is $500 per annum, A. spends $400, and B. ${ ()'. A man had two silver cups of unequal weight, hav- ing one cover to both, 5 oz.; now if the cover is put on the lesser cup, the whole will be double the weight of the greater cup. Again, if the cover be put on the greater cup, it will be 3 times as heavy a*s the lesser cup, what U the weight of each cup? Ans. 3 ozs. lesser — greater 4 ozs. 7. Divide 10 into 2 such parts, so that 9 times the les- ser number will be equal to 6 times the greater? Ans. 6 and 4. 8. The sum of 2 numbers is 50; now, if you divide the greater part by 7, and multiply the lesser by 3, the of the quotient and product will be equal to the giren bomber. Required the parts? Ans. 35 and 15. 9. A young gentleman having asked his father how old he was, received the following reply: seven years ago my age was in a three-fold ratio to yours; but, if we should both happen to live seven years hence, my age shall be just double that of yours. Required their several ages? 'Ans. 49 and 21 years. 238 [TIES. C/2 19 -< f R I. J ft O © s © 1 i t* h ' • tfi QD tt iH CO ^ £ J^ ^ oo — t — CI 00 PC — O 00 © 00 ~* 00 00 <N CO '-*. 00^ 00^ 00^ ^ r* T-. 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C — kfi X i ~ CO iT5 Q « — ~> - cc <^ co i> ci oo oo k © >c '0»-iHat>- 8 oO'-comoO'-ioomwa^i''. .-. «N — W re — .r. sc x r. - ^t •* m ^ r. - m _ ^ ^ _ _ _ c* i tH(NcoTjiiov0^aoiOH(Nn^o # >s i es. 239 ANNUITIES AT COMPOUND INTER I Definitions. — Annuity is a certain sum of money to be paid at regular periods, either for a limited time or for Present worth, or value of an annuity is that sum, which being improved at compound interest, will be suf- ficient to pay tin- annuity. The amount, of an annuity Is the compound interest of each payment added to their sum. To find the amount of an annuity at compound interest, we adopt the following. -Make $1 the first term of a geometrical series, and the amount of $1 at the given rate per cent, the ratio, carry the series to so many terms as the number of >, and find its sum. Multiply the sum thus found by the given annuity and the product will be the amount. CASE I. ample. — 1 . What will an annuity of $60 per annum, payable yearly, amount to in 4 years, at 6 per cent? 2 3 1x1.06x1.06+ 1.06 = 4.374.61, the tabular number, answering to 4 years, at 6 per cent. 4 Table 1, 1.06--1 1.06—1 X60 = $262.47c. 6m. Ans. 7461$X60 — *262.47c. 6m. as before. 2. What will an annuity of $30 amount to in 3 years? Ans. $95.50c. Sm. 3. What will an annuity of $200 amount to in 5 years at 6 j Ans. $1 127.41c. Sm. CASE II. When the payments are to be made half yearly or quar- teily, the amount for the given time, found as before, in table first, multiplied by the tabular number answering ei-o to the given rate per cent, for quarterly or half yearly payments, will be the true amount. TABLE SECOND. The construction of this table is from an Al- gebraic Theorem given by the learned Mons. De Moivre, in his Treatise of Annuities on Lives, which in words is thus: For half yearly payments take a unit from the ra- tio, and from the square root of the ratio, half the quotient of the first re- mainder, divided by the latter will be the tabular number; lor quarterly payments, use the fourth root, as above, and take one-fourth of the quotient. 1 . What will an annuity of $200 amount to in 5 years, to be paid in half yearly payments, at 6 per cent, per annum? Ans. $1144 8c. 2m. -f Agreeably to table first, the tabular number is 5.63709 X 200 = $1127 41c. 8w. X 1.014781, (tabular number answering to 6 per cent, in table second, for half yearly payments,) = $1144 08.2m. -f answer. J. What will all annuity of $500 amount to in 5 years, at 6 per cent.? Ans. $2818.54.6 -f. 3. What will an annuity of $1000, payable yearly, amount to in 10 years? Ans. $13180.79.4 -f. 4. What will annuity of $30, payable yearly, amount to in 3 years? Ans. $95.50.8 -f . ! 1 t ^L H 1 t* ^ p 3 1,0071 if) 1,011181 3i 1,00^ 1,019031 4 1,009902 1,011 1,011126 1,016720 5 1,012348 1,018559 54 1,02<> 1,0117^1 7 1,017204 »S80 ITIES. •J 11 g -s I ^ * Kfi fe w £ •— « M H 5: ~~ «2 D z s fc V < * 3 - - 8 — — — — .<ocooi~i*or^© C (N W 00 ~ " X CO it — . t*» QC O C 3> t*« OOO^Ol^OOOGCO^ ^O^'OWCOMWCMNM - , 2 ,r. <^ C5 O O* CO <* 'O t- to C0C0C0C0C0r$«Tf»"*Tj«-«*T$« ^^«?hONC)^4(7»^ CO <N t^ 't — < IOA60Q 8 o - o ^ oo w i- - - x : i (fly.- N ^> «© »~ k — • - <* h oi o o oi r — * *^ COOO^-COiOOOO'—COiOt^ «iftiRtnift50«C(0»0 J! t*00OiO^-C^C0'*»r5"«£>t» £ c<o<c<cocococococococo 8 -r. •noMW^h'fNoiooco 3 — r <--.'o-oO">*o»oooo c»t^ — -^oo»-Tj<t^oco»cr^o <£> C!ROOOHHi(N(NNO)n i 1 ' j ^oo cot-«~t<ococo-f<o*»n>rt -*iftO«)00(N-«ONrfi-*X -^O'tCOCN'-iOOCOCii-i C0O5h-^C5OC»^(?500a0C0»0 5, a>r>-coi^ooootoc»cooooi05t^ J56 co^ CO^ c^ ^ O^ ^ 00^ i-^ ^ l^ ©^ co^ lO a? cT cT -^ *-T of cf &f of cf crT r$T -^ J5 -* it. » i* x a o h o « ^ m -o t* | <r>co<N:o**-*-<'*o*t^»o~*co CiO>^©<O01C00JOC0t^Tj<C0 <j W«O-HKC0C3^»Q0000» P5W«»f5W^0»O«O«000(N ^ *> ■^.co^co^— oooocoooa *, Oi^ 00^ V^ r^ <N O^ lO C^ 00^ 00^ GO CO^ 00^ cf *■» of cf «* rjT uf <o «f i>" |C <£ & '-O00ONWP5rv - — - > ~ 6 oo^^m^cj^^fNm-io^ <j C M (N 00 ^ tJ. (N iO 1 O*C^C0iOC5»ri<OC0t>.»-COC0C0 — x <© © c* © «© o *i c^oq^i^if^c^o^t^^^t^ co go co id cT .-T cT c»f <*" io *f <©" i>^ i>" c<f aT of & rtWco^o^^oooiOHWw . — . __ _ ._ . 21 24-2 ANNUITIES. EXPLANATION OF THE TABLE. What is the present worth of $1, to continue for 4 years at 6 per cent, per annum? Ans. 3.465106, agree- ing with the tabular number opposite to 4 years at 6 per cent, per annum. First, find the present worth of $1, by discount for 1 year at 6 per ct. per annum, which is $0.943396 2d year the present worth is - 0.889996 3d " " " is - 0.839619 4th " " Ci is - 0.792094 $3.465105 Tabular number for 4 years at 6 per. cent, as in table 3d, 1 . What is the present worth of $50 per annum for 6 years at 6 per cent, per annum? Ans. $245.86c. 6m. CASE III. Annuities in Reversion. The annuity, time, and rate given, to find the present worth as in case 2. Multiply the number, under the rate 1 and opposite th etime in table 3d, by the annuity, the product will be the present worth for yearly payments* If the payments are to be made half-yearly, or quarterly, the present worth so found for yearly payments, must be multiplied by the proper number in table 2d. Q. What is meant by annuities in reversion? A. Sums of money, which are paid yearly for a limited period, but which do not commence till after the expira- tion of a given period, are called annuities in reversion. Given the time of reversion, time of continuance and rate per cent, to find the present worth of the reversion. Rule. — Take two numbers under the given rate in table 3, corresponding to the different periods of time, viz; time of reversion and time of continuance, and take the differ- ence between the tabular numbers, answering to the times as above mentioned, and multiply that difference by the ANNUITIES. M annuity, for the present worth annually, if the payments he 1 or quarterly, we must use table 2 as above stated. 1 . The reversion of a freehold is $60 per annum for 4> l, to commence 2 years hence, what is the present worth, allowing 6 per cent, for prompt payment. Illustration. — Time of continuance 6 years. Tabular number for 6 years at 6 per cent, found in table 3, is - - ' - L - - 4.917324 Time of reversion; 2 years tabular No. 1.833393 3.083931 3.083931x60 = $185.03c.5ro. X Answer. What is the present worth of a reversion of a lease of $100 per annum, to continue 10 years; but is not to commence till the end of 2 years at 6 per cent Ans. $655.04*. 3. What is the present worth of a reversion of a lease for $120 per annum to continue 9 years, but not to com- mence till the end of 4 years at 5 per cent, to the purcha- ser* Ans. $701. 71c. 4ot. PERPETUITIES. Annuities which continue for ever, are called perpetu- CASE IV- Given the Annuity and rate per cent, to find the present worth. IU i.e. — Divide the annuity by the ratio less 1, for the present worth. .Vote, — Table 2d must be resorted to, as in temporary annuities, when the payments are half-yearly or quarterly. 1. What ii the present worth of an annuity of $150 to continue for ever, allowing f) per cent, to the purchaser? '/*.— 1.05— 1 = .05)150.00 $3000 Ans. SMM COMPOUND INTEREST. ALLIGATION. What is an estate of $260 per annum, to continue for ever, worth in present money, allowing 6 per cent, to the purchaser? Ans. $4333.33.3 -f . A property in fee simple rents for $120 per annum, what is the present worth, allowing 5 per cent, to the purchaser? Ans. $2400. DISCOUNT BY COMPOUND INTEREST. The ratio in compound interest is the amount of $1 for one year, which is found thus; as 100 : 106 : : 1 = $1.06 is the amount of $1 for oney. mpk 1. — What is the present worth of 600 for 3 : hence at 6 per cent, compound interest 1.0b 3 *■ 1.191016)600( = $503.77 -f Ans. i is the amount of $503.77, in 3 years at 6 per cent.? Ans. $600. 3. What is the present worth of $520, due 5 years hence, at 6 per cent, compound interest. Here 520 = l,331225£)520(— 390.62 answer. 1.06* ALLIGATION, From the Latin (ad. to, and ligo to bind,) it being necessary in sundry cases to link or bind the quantities. I \ shall not omit the rule of Alligation, the object of which is to find the value of several things of the same kind of different values. The following examples will sufficiently demonstrate it. CASK I. When the quantities and rates of the simples are given to find the rate of a mixture compounded of these simples. Rule. — Find the value of each quantity, according to their respective costs, then divide the sum of the pro- .m.t.ic \ no*. 245 ducts "by the aggregate of the quantities, and the quotient ■will be the average value 6f each quantity. 1. A wine merchant bought several kinds of wine, viz: 130 bottles which cost him 10 cents each. " at lf> cents each. 231 « at 12 *« " 27 " at 20 " " 130atl0cts. - 1300 463)5737(12 39 " 15 « = 11 463 SM •• 12 '• = 52772 — •♦ 20 m = M0 1107 926 463 bottles cost 5737 cts. 1810 1389 4210 3867 2. A grocer has 4 lbs. of tea at 90 cents per lb., 8 lbs. at 75 cents, and 6 lbs. at 110 cents, to be mixed together, what will a pound of this mixture be worth? Ans. 90c. 3. A grocer has 2 cwt. of coffee at $25 per cwt.; 4 cwt. at $20.50 per cwt. and 7 cwt. at $18.62£ per cwt. which he will mix together, what will 1 cwt. of this mixture be worth? Ans. $20.18i. CASK II. When the prices of all the simples, the quantity of one of them, and the mean price of the whole mixture are given fa find the quantities of all the rest. Rule 1. — Place the mean rate and the several prices, link them and take their differences, as in the preceding CMC. J. As the difference of the same name with the quan- tity given is to the differenced respectively, so is the given quantity to the several required quantities. 21« 246 ALLIGATION. 1 . What quantity of cofFee at 20 cents, and at 16 cU. per lb. must be mixed with 35 lb. at 14 cents to make a mixture worth 18 cents per lb.? Mean rate i8 Then, as 2 : 35 : : 2 2 : 35 : : 6 2. How much tea at 86 cents, at 94 cents, and at 105 cents per lb. ought to be mixed with 6 lbs. at 75 cts. per lb. for a mixture, to sell at 92 cts. per lb.? Ans. 18 lbs. at $1.05, 51 lbs. at 94c, 39 lbs. at 86c. CASS III. When the prices of the several simples, the quantity to be compounded, and the mean price are given to find the quantity of each simple. Rule 1. — Link the several prices and take their differ- ences as before. 2d. As the sum of the differences is to the difference opposite each price, so is the quantity to be compounded to the quantity required. 1. A grocer has three sorts of sugar, viz: 10, 11, and 8 cents per lb. how much of each sort must he take? 4 H 1 +2 =3 Mean rate 9 ] 10 = 1 =1 f 11- " = 1 = 1 Sum of differences, As 5 : 3 : : 40 = 24 at 8 ce 5 : 1 : : 40 = 8 at 10 5 : 1 : : 40= 8 at 11 5 ;nts. u 2. A vintner has wine at 130 cts. at 160 cts. and at 180 cts. per gallon, and he would have 32 gallons worth 145 cents per gallon, I demand how much of each sort he must have? Ans. 20 gals, at $1.30, 6 gals, at $1.60 and 6 gals, at $1.80. ARITHMETICAL PROGRESSI<. 24-7 ARITHMETICAL PROGRESSION. ;i a series of numbers or quantities increase or de- crease by a constant difference, it is called Arithmetical progression; as, 1, 2, 3, 4, 5 6; 1, 3, 5, 7, 9, 11; 6, 5, 4, . 1; 11, 9, 7, 5, 3, I. There are five things to be par- ticularly attended to in Arithmetical Progression; the first term, the last term, the number of terms, the com- mon difference, and the sum of all the terms. CASE I. The first term, common difference, and number of terms being given to find the last term, and sum of all the terms. Rule 1. — Multiply the number of terms, less one, by the common difference, and to that product, add the first tt'rm, the sum is the last term. 2. Add the first and last terms together, and multiply the sum by the number of terms, and half the product will be the sum of all the terms. 1 . A person sold 40 yards of muslin at 2 cents for the yard, 4 cents for the second, increasing 2 cents every yard, what did they amount to? Ans. $16.40. OPERATIOV. Nos. of terms 40 — 1 wm 39 x 2 = 78 + 2 = 80 last term. 1st term or extreme 2 Last do. or second extreme 80 82x40=* s y° =$16.40 2. A butcher bought 75 sheep, and gave 6 cents for the first, 8 for the second, &c> what did he give for the last, and what did the whole number cost him? Ans. For the last $1.54, the whole $60. :*. A travels uniformly at the rate of 6 miles an hour, and sets off upon his journey 3 hours and 20 minutes be- fore B; B follows him at the rate of 5 miles the first hour, 6 the second, 7 the third, and so on. In how many hours will B overtake A? Ans. 8 hour*. 248 GEOMETRICAL PROGRESSION. CASE II. U7ien the first arid last terms (or two extremes) are given to find the common difference. Rule. — Divide the difference of the extremes by the number of terms less 1; the quotient will be the common difference. 1. If the ages of 12 persons are equally different, the youngest is 18 years and the eldest 40, what is the com- mon difference of their age? Ans. 2 common difference. Illustration of the above question. 40 12—1 = ll)22f2 common difference. 2. The extremes are 3 and 45, and the number of terms is 22, what is the common difference? Ans. 2. 3. A man received "charity" from 10 different persons, the first 4 cents, the last 49 cents, what was the common difference, and what did the man receive? Ans. he received $2.65; com. dif. 5 cts. 4. The extremes are 3 and 39, and the sum of the series 399; what is the common difference? Ans. 2. GEOMETRICAL PROGRESSION. Geometrical Progression is the increase of a series of numbers by a common multiplier, or decrease by a com- mon divisor; as 2, 4, 8, 16, 32; 32, 16, 8, 4, 2; the ratio is the number by which the series increases or decreases. CASE I. To find the last term and sum of the series. Rule. — Raise the ratio to the power whose index is 1 less than the number of terms given. PROGRESSION. 249 Itiply the product by the first term, and th< suit will bfl the last term. 3. M tiltipl v the last term by the ratio; from the product subtract the first term, and divide the remainder by the 1 , for the sum of the sei i 1 . Ill buy 16 cords of wood, and agree to pay 2 cents for Che first. 1 for the second, 8 for the third, &c., doubling the price to the last, what will it cost me? uh. I 1,2, 3,4, ratio 2, 4, 8, 16 fourth power 16' 96 te •\ 8th power, fourth power 16 1536 256 4096 12th power. third power 8 3-2768 15th power, first term 2 S26 last term. 2 131072 2 first term. Ratio 2—1 = 1)131070 : 1 0.70 Answer. A person at the birth °t" his s °n> deposited in bank 1 cent, towards his fortune, promising to double it at the return of every birthday, until he was 21 years of age, what was his portion? Ans. $20,971.51. DUODECIMALS. 3. A gentleman consented to have his daughter married on New Year's day, and agreed to give her one dollar to- wards her portion, promising to double it, on the first day of every month for one year, what was her portion? Ans. $4095. 4. A thresher wrought 20 days and received for the first day's labour 4 grains of wheat; for the second 12; for the third 36, &c , how much did his wages amount to, al- lowing 7680 grains to make a pint, and the whole to be disposed of at $1 per bushel? Ans. $14187. 5. A sum of money is to be divided among 10 persons, the first is to have $10, the second $30 and so on in three fold proportion; what will the last have? Ans. $196830. DUODECIMALS. This rule is of great use to carpenters, joiners, &c. The name is derived from the latin words duo, 12 and decern, 10, and as the ratio is 12, it may with propriety be termed Duodecimals. As the French and other European nations divide their inch into 12 equal line6, so our American artificers, suppose the inch to be divided as follows: DENOMINATIONS OF DUODECIMALS. 12 fourths make 1 third, fourths marked thus "" 12 thirds make 1 second, thirds " 12 seconds u 1 inch, seconds " 12 inches " 1 foot, inches " I Feet " F EXAMPLES IN ADDITION. The ratio being 12 the rule is evident. F. I. " ■" F. I. " Add 642.11.8.8 Add 6464.10.9 436.10.6.4 4243. 9.4 Add 6468.10.5 Add 9346. 7.3 4864. 9.3 4842. 9.4 6968. 8.7 3464.11.6 DUODECIMALS. Mil ,X tFLIOATIONi Observe the following rules. multiplied by feet produce feet. Fri t multiplied by inches produce inches. mltiplied by seconds produce seconds. Inches multiplied by inches produce seconds. Inches multiplied by seconds produce thirds. Seconds multiplied by seconds produce fourths. 1. Multiply 5 feet G inches by 2 feet 4 inches. W I. 5.6 2.4 11.0 1.10.0 12.10.0 2. Multiply 8 feet 6 inches by 14 feet 9 inches. Ans. 125 ft. 4- in. 6 s. CASE II. Rule. — Multiply by the component parts, as in com- pound multiplication, and take parts for the inches as in practice. 1. Multiply 208 feet 8 inches 4 seconds, by 24 feet 3 inches 9 seconds. F. 1. M Operation. 208.8.4 0X4 1252.2.0 4 5008.8.0 3 in. is | 52.2.1 9 » is i 1H.0.6.3 5073.10.7.3 Answer. 2. Multiply 4 feet 7 inches by 6 feet 4 inches. Ans. 29 ft 0. in 4 s* 3. How many square feet are in a floor 48 ft. 6' in. longf by 24 ft. 3 in. broad? Ans. 1176 ft. 1 in. 6 s. MEXSURATIi 4. How many cubic feet of stone work are contained in 9 walls, each 30 ft. b in. long, 9 ft. 8 in. high and 2 ft. thick? Ans. 5307 ft. 5. How many cubic feet in a cellar 30 ft. 3 in. loo ft. 6 in. broad, and 8 ft. deep? Ans. 6(>f)5 ft. 6. What is the content of a marble slab 6 ft. 7 in. long, 2 ft. broad, and 1 ft. 9 in. thick? Ans. 23 ft. Oin. I 7. What is the superficial content of a stone 4 ft. 9 in. long, and 3 ft. 9 J in. broad? Ans. 18 ft. in. 1 s. 0" t. MENSURATION. IMPORTANT TO SHIP BUILDERS. A concise Ride to find the length of M< Rule 1. — Multiply the length of the keel by 2, and divide the product by 3, and then to the quotient add the breadth of the beam, and the sum will be the length of the main m 1. Suppose a ship to be 84 feet by the keel, and 31 it. by the beam, what is the length of her mast? Ans. 87 ft. 2. Suppose a ship to be 108 feet by the keel, and 40 feet by the beam, what is the length of her main mast? Ans. 112 feet. ANOTHER METHOD. When the length and thickness of Masts is required in yards. Rule. — Add the breadth of the beam and the depth of the hold in feet together, and divide the sum by 1.5, and the quotient will be the length of the main mast in yards. 1. Admit a ship whose keel in length is 73 feet, and the breadth of the beam 2S.5 feet, and the depth of the hold 12 feet, what is the length of her main mast? Feet. Ans. 81 feet. Breadth of the beam, 28.5 Depth of the hold, 12.0 1.5)40.5(27 yards or 81 feet. MENSURATION. Now to find the thickness, it is customary to allow i of the lenirtli in fbet for the thickness in inches; in that case, a main mast 81 feet long, must be 27 inches thick. If a ship of 100 tons be 44 feet long at the keel, of what length must the keel of a ship be that carries 220 tons? tans. tons. Say as 100 : 220 : : 44 s = (85184) 187404.80, whose cube root is 57.226, the length of the keel sought To find Ship's Tonnage by Carpenter's Measure. Rule. — For single-decked vessels, multiply the length and breadth at the main beam, and depth of the hold together, and divide the product by 95. For double-decked vessels, take half the breadth of the main beam, and work as above directed. 1. What is the tonnage of a single decked vessel, whose length is 60 feet, breadth 20, and depth 8 feet? Ans. 101 y'y tons. Operation.— 60 x 20 X 8 = 101 T V tons. 95 2. What is the tonnage of a double-decked vessel, whose length is 65 feet, breadth 21 feet 6 inches, and depth 10 feet 9 inches? Ans. 158*t tons. TO FIND GOVERNMENT TONNAGE. Government Rule. — "If the vessel be double-deck* ed, take the length thereof from the fore part of the main stem to the after part of the stern-post above the upper deck; the breadth thereof at the broadest part above the main wales, half of which breadth shall be accounted the depth of such vessel, and then deduct from the length | of the breadth; multiply the remainder by the breadth, and the product by the depth, and divide this last pro- duct by 95, the quotient whereof shall be deemed the true contents or tonnage of such ship or vessel; and if such ship or vessel be single-decked, take the length and MENSURATION, breadth, as above directed, deduct from the said 1. $ of the breadth, and take the depth from the undci of the deck-plank to the ceiling in the hold, and the multiply and divide as aforesaid, and the quotient shall be deemed the tonnage." 1 . What is the government tonnage of a single-decked vessel, whose length is 69 feet 6 inches, breadth 22 feet 6 inches, and depth 8 feet 6 inches? F. I. F. 1. Illustration of the rule 69.6 breadth 22.6 Deduct 13.6 3 56.0 5)67.6 Breadth 22.6 13.6 112.0 119 1232.0 6 in. is i 28.0 •/. ,4 feet tons. feet 1260 x 8.6 = ' Vi 1 ° = H2f| 2i Required the tonnage of a single decked vessel, by carpenter's measure, whose length is 70 feet 6 inches, breadth 24 feet 8 inches, and depth 9 feet 10 inches? Ans. 180 tons. A. What is the Government tonnage of a double-deck- ed vessel of the following dimensions: length 82 feet 3 inches, bieadth 24 feet 3 inches, and depth 12 feet 1£ inches? Ans. 209$ f tons. BOARD OR LUMBER MEASURE. Rule* — Multiply the length in feet by its breadth in inches, and divide by 12 for the content. 1. What is the content of a board 24 feet long and 8 inches wide? Ans. 16 ft. 2. What is the content of a board 30 feet by 16 inches? Ans. 40. MENSURATH' 255 \ hat is the content of a board 14 feet by 15 inch Ans. 17J ft. 1. What il the content of a board 18 feet by 15 inch. Ans. 22* ft. TO MEASURE SCANTLING OR JOIST. Rule. — IMultiply the depth and width taken in inches he length in feet, divide the product by 12, and the quota nt is the content in feet. 1 . How many feet are there in 3 joist, each of which 1 5 feet long, 5 inches wide and 3 inches thick? Ans. 56i feet. 2. How man^ feet in 20 joist 10 feet long 6 inches wide, and 2 inches thick? Ans. 200 feet. CASE II. When a Board is wider at one end than the other. General Rule. — Take the breadth in the middle, or add the measure of both ends together, and take the sum for a mean breadth which multiply by the length for the content. 1. Suppose a board be 10 feet long and 10 inches wide at one end, and 34- inches wide at the other end, what is its superficial content? Ans. IS 3 feet. Illustration, :i L 10 » J sum, 44 22 medium or mean breadth. 1 1 1 feet long 12)220 18 3 feet answer; or, if the length be • t and inches, reduce the length to inches, which being multiplied by the mean breadth in inches and divided by 1 14, we get the content hi feet. 256 MENSURATION. Q. Why do we divide by 144? A. Because, when we multiply inches by inches, the product is square inches; therefore, we divide by 14 1, 144 square inches being = to 1 square foot. PlPEHOia ROOMS. There is a room papered, the compass of which is 47 feet 3 inches, and the height 7 feet 6 inches, what is the content in square yards? Ans. 39^ yds. CARPENTER'S WORK. Roofing, flooring, partitioning, and the principal car- pentry in modern buildings, are measured by the square oi 10 feet, that is 100 feet. Rule for roofing. — Multiply the depth and half depth by the front, or, the front and half front by the depth, and you get the content. The dimensions are ta- ken in feet and inches. I . I fa floor be 49 feet 6 inches long, and 26 feet 6 inches broad, how many square feet? Operation. 49.5 2475 2970 990 1311(75 Ans. 13,11 ft. 9 in. or 13s. 1 If. 9i. 12 9)00 BRICKLAYER'S WORK. Bricklayers are generally paid by the day or perch. 1. Suppose a garden wall to be 254 feet round and 12 feet 7 inches high and 3 bricks thick, how many square rods does it contain? Ans. 23 £ sq. rods. F F I 254+12.7 '= 3196.2x2 - 6392.4 =23^ sq. rods, 272 MENSURATION. 257 >/<?.— As the standard thickness is 1 \ brick thick, then 16 J feet long, 1 foot thick and 1 J feet high *= 24.75 feet, ;>erch, hence we multiply the length, breadth and thickness of the wall together, and divide by 24.75 for :. umber of perches required. DIGGING. Cellars, vaults, clay for brick, cunuls, &rc, are measured by t//t solid I/ in! of J 7 feet. Uui.k. — Multiply tlu- length, width and depth, together and divide the product by 27, for the number of cubic How many yards of digging in a cellar 25 feet long, 20 feet wide and 10 feet 6 inches deep? Ans. 194f cub. yds. ■'•ft. — A solid yard of clay will make 7 or 800 brick, and 3.} bushels of lime and half a load of sand will be sufficient to lay 1000 brick. To find how many thousand brick will be required for building a house of any given dimensions. Suppose a house of the following dimensions, viz: 84 long, 40 feet wide, 20 feet high and the walls to be 1 foot t: Ans. 105,408 bricks. Rile. — Deduct the thickness of the wall, from the length of each side, because the inner sides are 1 footless in height than the outer sides. This rule is unquestion- . correct. OPERATION. 84—1 — 83 feet and 40—1 « 39 Now 83 X 2 a=» 166 sum of 2 sides in length, and 39 X 2 = 78 sum of 2 do. in breadth, 244 compass 20 feet in height. 4880 X 1728 = 8432640 cub. in. \uw allowing a brick to be 8 inches long, 4 inches 258 MENSURATION wide, and 2\ inches thick, there would be 80 cubic inches in a brick, hence 84.3264(0 «= 105,408 brick, Ans. 8(0 J. How many thousand brick 8 inches long, 4 inches wide, and 2\ inches thick, will build a wall in front of a church which is to be in compass 240 feet long, 6 feet high and 1 foot 6 inches wide? Ans. 51.840 bricks. 3. How many shingles will it take to cover the roof of a barn 40 feet long, allowing the length of the rafters to be 16 feet 6 inches, and 6 shingles to cover 1 square footj what will they cost at $1.25 per 1000? Ans. 7,920 shingles; cost $9.90. MENSURATION OF SUPERFICES AND SOLIDS. n \) PROBLEM 1. A To find (he area of a square. Rule. — Square the side: and the pro- duct or rectangle will be the superficial C content. 1. A lot of ground is 10 perches square, what is the area? problem 2. Ans. 100 ps. = 2 r. 20 p. JVbfe. — A square is a parallelo- A gram, but a parallelogram is not a u square, because it is an oblong ~> whose length and breadth are un- equal. C D 2. What is the content of a board 15 feet long and 2 feet wide? Ans. 30 feet 3. What is the difference between a floor 40 feet square and 2 others, each 20 feet square? Ans. 800 feet. 4. There is a square of 3600 yards area; what is the side of a square, and the breadth of a walk along each side of the square at each end, which may take up just one half the square? Ans. 42.42 -f yds. side of the sq. 8.78 + yds. breadth of walk. MENSURATION. 259 a PROBLEM 3. A B To find the area of a rhomb \{\ i.k. — Multiply the length of Ihe base by the perpendicular height. 5. The base of a rhombus is 14 Q feet C and its height 6 feet, required — D the area? Ans. 84 feet. problem 4. To find the area of a triangle. 1 1 le. — Multiply half the base by the perpendicular height, or if the perpendicular is not given, A add the three sides together, take half that sum, subtract each side severally; from the half mm multiply the half sum and the three differences to- - r, and the square root of the product will be the area. 1. Required the area of a right angled triangle whose base is 40 and perpendicular 30 perches? Ans. 600. 2, Required the area of a triangle whose sides are 10, 12 and 18 perches respectively? Ans. 56.57 perches. PROBLEM 5. CASE I. By having thr diameter of a circle to find the area. Rule. — Square the diameter, and multiply the product by .7854 for the area. CASE II. By having the circumference of a circle to find the area. I. — Square the circumference and multiply that square by .07!' 1 . The diameter of a circle is '24, required the area? Ans. 452.4904. i'iie circumference of a circle is 80, required the area? Ans. 509.312. 260 MENSURATION. CASE III. By having the diameter to find the circumference. Rule — Multiply the diameter by 3.1416, and you get the circumference. It* the diameter of a circle be 24, what is the circum- ference? Ans. 75.3984. CASE IV. By having the circumference of a cube to find the diamtft r. Rule. — Multiply the circumference by .31831 and the product is the diameter. PROBLEM 6. I have a circular field 50 rods in diameter, what is the .-iilc of a square field, that shall contain the same area? Ans. 44.31. solids Are figures having length breadth and thickness. PROBLEM 7. To find the content of a cube or paral- Q lelopipedon, whose side is 18 inches. Rule. — Multiply the length, height, and breadth continually together, and the product is the content. A NJ Jb 1. How many cubic feet in a cube whose side is 18 inches? Ans. 3| feet. '2. What is the content of a parallelopipedon whose length is 6 feet, height 2 feet, breadth 1 \ feet? Ans 18ft, 3. A cellar is 50 feet long, 38 feet wide and 12 feet deep, how many cubic yards of earth has been taken out in digging, and what was the expense of digging it at 10 <;ts. per cubic yard? Ans, 844.44 + cubic yds.; expense $84,45 nearly. O MAN SITUATION. 261 PROBLEM 8. To find the solidity of a Prism. A| B. — Multiply the area of a base or end UVquired the solidity of a triangular prism whose length is 10 {eel, and the three sides of lingular base are 5, 4 and 3 ft. Ans. 60ft. CI PROBLEM 9. To find the solidity of a Cylinder. It, — Multiply the area of the base by the length. The diameter of the base of a cylinder is 10 inches and its length 24 feet, required the solidity? Ans. 13.09 feet. PROBLEM 10. To find the solidity of a cone or pyramid. Rule. — Multiply the area of the base by £ of its height. & What is the solid content of a cone whose heighth feet, and the diameter of the base 2£ feet? Ans. 20.45 + feet. PROBLEM 11. C To find the svperfrcs of a Cone. Kri.n. — Multiply the circumference of the base by half its slant height. 7. What is the convex surface of a cone, whose slant height is is 20 (eet, and the cir- cumference of its base 9 : Ans. 90 feet. A PRor.r.r.M 12. To find the solidity of t m of h cone or pyramid. \{i LB. — Multiply the di. . ters of the two bases togeth- er, and to the product add f the square of the difference of the diameters; then multiply this sum by .7854- and GUAGING. the product will be the mean area between the two bases; Lastly, multiply the mean area by the Frustum ef a coae. length of the frustum, and the product will be the solid content. 8. What is the content of a stick of timber whose length is 40 feet, the diameter of the larger end 24 inches, and the smaller end 12 inches? Ans. 73£ ft. nearly. PROBLEM 13. To find the solidity of a sphere or globe. Rule. — Multiply the cube of the diameter by .5236. 9. What is the solidity of a sphere or globe, whose axis or diameter is 12 inches? Ans. 904.78 + inches. PROBLEM 1 1. To find the convex surface of a sphere or globe. Rule. — Multiply the diameter by the circumference. 10. Required the superficial content of a globe whose diameter is 24 inches? Ans. 1809.55 -j- inches. To find how large a cube may be ait from any given sphere, or be i in it. lit le. — Square the diameter of the sphere, divide that product by 3, and extract the square root of the quotient for the answer. 1. How large a cube may be inscribed in a sphere 40 inches in diameter* Ans. 23.09 -f inches. GUAGING. The business of cask guaging is Caako^h^jmjmm^ commonly performed by two in- £ BBNk ' struments, namely, the guaging or sliding rule, and the guaging or jdiagonal rod. >G3 To find the solid content of ( r..— Kind the mean diam- by taking the mean of the bilge and head diain- « tar, then mul- tiply the mean jg tabu:. -:-' a "3 00 6-2 *5 ,o o 6 -° 5 O o> . B •a u 1 fcX) S5 § ••* o ^ JC 3 1 1 h- §i<2 g diameter in inches by the length of the cask in inches. And, again, multiply this product by the mean diameter; deduct one-fifth of the sum so found for the roundness of the cask, and reduce the remainder to feet and inches, by the rule for measurement, which is correct if the con- tent be required in solid feet, or, divide by 232 for ale :is. xpU 1. — If a cask be in length 3 feet 9 inches, its head diameter 2 feet 6 inches, and its bung diameter 2 10 inches, what are its solid contents in ale gallons? Length 3 ft. 9 in. Multiply 45 length. 12 32 45 90 Head diameter, Bung diameter. 135 2 ft. 6 in. 2 ft. 10 in. 12 12 1440 — 32 Again to be 30 Bung diam. 34 inches multiplied. Head « 30 -Jsso 4)64 4320 .46080 Mean diam. 32 in. Deduct j 9216 gals. qts. 282)36864(130 3 2. Each side of the square base of a vessel is 40 in- ches, and its depth 10 inches. Required its contents in ale gallons? Ans. 56.7 gallons. ABSTRACT OF MECHANICS. 8. The diameter of a cylindrical vessel is 32 inches, and its internal depth 4-5.5 inches. Required its content in ale gallons? Ans. 129.78 gals. 4. How many bushels of grain will a box contain that is 15 feet long, 5 feet wide, and 7 feet high? Ans. 421.8 bushels. ABSTRACT OF MECHANICS. OF MATTER. Matter, is possessed of the following properties, viz: ///y, extension, </i risibility, mob it «, ut tract ion and repulsion. 1. Solidity is that property by which two bodies can- not occupy the same place at the same time. It is some- times called the impenetrability of matter. 2. Extension, like the solidity of matter, is proved by the impossibility of two bodies co-existing in the same place. 3. Divisibility, is that property by which bodies are capable of being divided into parts removable from each other. 4. Mobility expresses the capacity of matter to be moved from one position or part of space to another. 5. Inertia, designates the passiveness of niatter, which, if at rest, will forever remain in that state until compelled by some cause to move; and, on the contrary, if in mo- tion, that motion will not cease, or abate, or change its direction unless the body be resisted. 1. Space is either absolute or relative. 2. Absolute Space, is merely extension, immoveable, illimitable and without parts, yet it is usually spoken of as if it had parts. Hence the expression: g Relative Space, signifies that part of absolute space occupied by any body, as compared with any part occu- pied by another body. ABSTRACT OF MECHANICS. ATTRACTION. 4. Attraction denotes the property which bodies have to approach to each otl. There are five kinds of Attraction: 1, the attraction of cohesion; 2, of gravitation; 3, of electricity; 4, of mag- netism; 5, of chemical attraction, a is exerted only at very small distances, its strength varies in different kinds of matter, and is suppo- to be the cause of the relative degrees of hardness of different bodies. fiOutry .Ittraction is only a particular modification, or branch of the attraction of cohesion. , Gravitation decreases from the surface of the earth vj)wan/\, as the square of the distance increases; but from the surface of the earth downwards, it decreases only in a direct ratio to the distance from the centre. REPULSION. 1. Repulsion is that property in bodies, wheieby, if they are placed just beyond the sphere of each other's at- traction of cohesion, they mutually fly from each other. 2. Oil refuses to mix with water, from the repul- sion between the particles of the two substances; and from the same cause, a needle gently laid upon water will iwim* MOTION. Absolute Motion, is the actual motion that bodies have, considered independently of each other, and only with regard to the parts of space. lidnhvr motion, is the degree and direction of the mo- tion of one body, when compared with that of another. decelerated motion, is understood, when its velocity continually increases. Retarded motion, when the velocity continually de- creases, and the motion is said to be uniformly retarded. H hi'ii it decreases equally in equal times. Uniform motion is estimated by the time employed in 23 266 ABSTRACT OF MECHANICS. moving over a certain space, or in other words, by the space moved over in a en tain time* To ascertain the vrloritij. Divide the space run over by the time. CASE II. To ascertain tlu <// stance. Multiply the velocity by the time. In accelerated, motion the space run over (or distance) is as the square of the time. 1. A body acted upon only by one force, will always move in a straight line. 2. Bodies acted upon by two single impulses, whether equal or unequal, will also desenbe a right line. The Momentum of a body is the force with which it moves, and is in proportion to the weight or quantity of matter, multiplied into its velocity. The action of bodies on each other are always equal, and exert in opposite directions; so that any body acting upon another, loses as much force as it communicates. CENTRAL FORCES. The central forces are its centrifugal and centripetal forces. The centrifugal force is the tendency which bodies that revolve round a centre, have to fly from it in a tangent to the curve they move in, as a stone from a sling. 2. The centripetal force is that which prevents a body from flying off by impelling it towards the centre, as the attrac- tion of gravitation. CENTRE OF GRAVITY. 1. The centre of gravity is that point in a body about which all its parts exactly balance each other in every direction. 1. A vertical line passing through the centre of gravity of a body, is called the line of direction. MECHANICAL POWERS. 267 \Vh< n the line of direction falls within the base of a body, that body cannot descend; but if it falls without ! the base, the body will fall. F~l MECHANICAL POWERS. That body which communicates motion to another is called the power. The body which receives motion from another, is called the weight. The mechanical powers are five the lever, the wheel or .axle, the pul/y, the screw and the wedge, to these, may be added the inclined plane. OF LEVERS. There are 3 orders or varieties of levers, wherein the weights, props or moving powers, may be differently applied to the vectis or inflexible bar, in order to effect mechanical operations in a convenient manner. A lever is said to be of the first order when the prop is between the weight and the power; of the second order, whin the weight is between the prop and the power; of the third order, when the power is between the prop and the weight. A power and weight acting upon the arms of a lever will balance each other, when the distance of the point at which power is applied to the lever from the prop, is to the distance of the point at which weight is applied, as the weight is to the power. To find what weight may be raised by a given pov r. B -As the distance between the body to be raised or balanced, and the fulcrum or prop, is to the distance between the prop and the point, where the power is ap- plied, so i.s the power to the weight which it will balance. J . If a lever be 100 inches long, what weight lying 7$ 268 UANICAL POWERS. inches from the end, resting on a pavement, may be moved with the force of 168 lbs. lifting at the other end of the lever? Ans. 2072 lbs. 2. A water wheel turns a crank working three pump rods, fixed 6 ftet from the joint or pin which their seve- ral levers, each 9 feet in length are fastened on account of the intended motion at one end, the suckers of the pumps being put in operation by the other, proves them to be levers of the third order. Now if the crank in this case plays 9 inches round its centre, what is the length of the stroke in each of the barrels? Ans. 27 inches. 'A. With what force of water ought that water wheel be driven, (circumstanced as in the last question) which raises 3 cubic feet of water at every revolution of the wheel, each experimentally weighing 62* lbs. Avoirdu- pois, the friction of the wheel rejected? Ans. 281 i lbs. 4. At what distance from a weight of 1530 lbs. must a prop be placed, so that a power of 170 pounds, applied 9 feet from the prop, may balance it? Ans. 1 foot. THE AXLE OR WHEEL. The wheel or axle is a wheel turning round together with its axis the power is applied to the circumference of the wheel, and the weight to that of the by means of cords, an equili- brium is produced in the wheel and axis, when the wheel is to the power, as the diame- ter of the wheel to the diameter of its axis. Rule. — As the diameter of the wheel is to the diame- ter of the axle, so is the weight to be raised by the axle, to the power that must be applied to the wheel. CASE 1. 1. If the diameter of the axle be 6 inches, and the diameter of the wheel 4< feet^ what power must be applied to the wheel to raise 960 lbs. at the axle? Ans. 120 lbs. M1CHAWICAL POWERS. 269 CASE II. 2. If the diameter of the axle be 6 inches, and the diameter of the wheel 5 feet, what power must be appli- ed to the axle to raise 300 lbs. at the wheel? Ans. 24001bs. CASE III. i the diameter of the axle be 8 inches, and 300 lbs. applied to the wheel to raise 24-00 lbs. at the axle, what is the diameter of the wheel? Ans. 64- inches. CASE IV. : the diameter of the wheel be 64- inches and 300 lbs. applied to the wheel to raise 2400 lbs. at the axle, what is the diameter of the axis? Ans. 8 inches. PULLEY. The pulley is a small wheel moveable about its axis by means of a cord which passes over it. When the axis of the pully is fixed, the pully only changes the direction of the power; If moveable pulleys are used an equilibrium is produced when the power is to the weight, as one to the number of ropes applied to them, if each moveable pully has its own rope, each pul- 11 be double the power. To find the weight that may be raised by a given power. Rule. — Multiply the power by twice the num- ber of moveable pulleys, and the product is the weight. CASE L 1. What power must be applied to a rope that passes over one moveable pully, to balance a weight of 4-00 lbs.? Ans. 200 lbs. CASE II. J. What weight will be balanced by a power of 20 lbs. attached to a cord that passes over three moveable pulleys! W - Ans. 120 lbs. MECHANICAL POWERS. CASE III. 3. What weight will be balanced by a power of 100 lbs. attached to a cord, that passes over 2 moveable pulleys? Ans. 400 lbs. CASE IV. 4. If a cord that passes over two moveable pulleys be attached to an axle 6 inches in diameter, and if the wheel be 60 inches in diameter, what weight may be raised by the pulley, by applying 100 pounds to the wheel] Ans. 4000 lbs. INCLINED PLANE. An inclined plane, is a plane which makes an acute angle with the horizon. The motion of a body descending down an inclined plane is uniformly accelerated. The force with which a body descends by the force of attraction down an inclined plane is to that, with which it would descend freely as the ele- vation of the plane is to its length, or as the size of its angle of inclination to radius. To find the power thai will draw a weight of an. inclined plane. Rule. — Multiply the weight by the perpendicular height of the plane, and divide the product by the length. 1. An inclined plane is 60 feet in length, and 15 feet perpendicular height, what power is sufficient to draw up a weight of 1000 lbs. Ans. 250 lbs. 2. A certain Railroad, 1 mile in length has a perpendi- cular elevation of 20 feet, what power is sufficient to draw a train of baggage cars, weighing 79200 lbs. up this elevation? Ans. 300 lbs. THE SCREW. The screw is a cylinder which has either a prominent part, or a hollow line passing round it in a spiral form, so inserted in one of the opposite kind, that it may be raised or depressed at pleasure, with the weight upon its up- MECHANICAL POWERS. 271 per or suspended beneath its lower surface. In the screw the equilibrium will be produced, win in the power is to listance between the two contiguous threads, in a direction parallel to the axis of the screw to the circumference of the circle described by the power in one revolution. To find the power that should be applied to raise a giiu a ■\ r hf. Rule. — As the distance between the threads of the screw is to the circumference of the circle described by the power, so is the power to the weight to be raised. Note. — One-third of the power is lost in overcoming friction. 1. If the threads of a screw be 1 inch apart, and a power of 100 lbs. be applied to the end of a lever 10 feet long, what force will be exerted at the end of the screw? Ans. 75398.20 + lbs. THE WEDGE. The wedge is composed of two inclined planes whose bases are joined. When the resist- ing forces and the power which acts on the wedge are in equili- F ^ E brio, the weight will be to the powe^ as the height of the wedge to a line drawn from the middle of the base to one eide, and parallel to the direction in which the resisting force acts on that side. To find the force of the Wedge. Rule. — As the breadth or thickness of the head of the wedge is to one of its slanting sides, so is the power which acts against its head to the force produced at its side. 1. Suppose 100 pounds to be applied to the head of a wedge that was 2 inches broad, and 20 inches long, what force would be effected on each side? Ans. 1000 lbs. 272 STEAM ENGINE. ON THE STEAM ENGINE. Steam at the temperature of 212° is 1800 times its bulk in water; or, one cubic foot of steam, when its elasticity is equal to thirty inches of mercury, contains one cubic inch of water. Therefore, when an engine in good order, is performing its regular work, the effective pressure may be taken at eight pounds on each square inch of the sur- face of the piston. CASE I. To calculate the power of an Engine. It has been demonstrated in the Franklin Institute, of Philadelphia, that a horse can draw 200 lbs. at the rate of 2 J miles an hour, or 220 feet in a minute with a continuance drawing ever a pully, that is, 200 x 220 = 44000 lbs. at 1 foot per minute, or 1 lb. at 44000 feet per minute. Rule 1.— Multiply the area of the cylinder by the effective pressure, say 8 lbs., the product is the weight the engine can raise. Multiply this weight by the num- ber of feet the piston travels in one minute, the product will give the momentum, divide this momentum by a horse power, and the quotient will be the number of horse power in the engine. 2. The velocity of an engine being 220 feet per minute, 25 inches of the area of the cylinder is equal to one horse power. TABLE FIRST. Length of Number of stroke, strokes Feet 2 " 3 a a a a n 43 32 25 21 19 17 15 14 Feet per minute. 172 192 200 210 228 238 240 250 <* 273 1 . What is the power of an engine, the cylinder being aeter and stroke 5 feet? 2 = 74.44 horse power. 44000 y % What size of cylinder will a 60 horse power engine require when the stroke is 6 feet? ^°°J^9 = H47.39 + area of cylinder. .s x 8 Examples calculated by Rule Second. 1 . What diameter is the cylinder of a 40 horse engine, common pressure? v/40x 25 .7854 35.7 say 35£ inches diameter. CASE II. To find the power to lift a weight at any velocity. Rule. — Multiply the weight in pounds, by the velo- city in feet, and divide by the horse power, the quotient will be the number of horse power. TABLE SECOND. 3g~.». ~~~~~ «~..w~.»~.~ .. ..-..»«;% N the <J)'cc! : The area equal to one on each inch of }>i*t<>n Ii horse power will be pounds. 3.70 inches. 48 ' " 4.17 " 43 " 4.65 " 38 " ; « 33 « 6.06 " 28 " 7.14 " 23 « 8.70 " 18 " 11.11 13 « 15.46 " 8 " 25.00 " \i x 1 . What diameter is the cylinder of a 40 horse engine, .fsure 33 lbs. on the square inch? V 40 X 6.06 .7S54 say 17 a inches diameter. 274 i:ral tiieor! J. The cylinder of an engine is 40 inches diameter, and the effective pressure is 20 lbs. on the square inch, what is the power of the engine? 40* x .7854 = 1256.64 1256.64 9a *~>* = 157.0a horse power. GENERAL' THEOREMS/ THEOREM 1. ^Vhen the sum and difference of any two numbers are given, to find the numbers. Rule. — To half the sum add half the difference, for the greater number. From half the sum, take half the differ- ence for the lesser number. theorem 2. The product of the sum and difference of any two num- bers is equal to the difference of their squares. theorem 3. If the difference of the squares of two numbers be divi- ded by their difference, the quotient will be the sum; and if by the sum of the numbers, the quotient will be the difference. theorem 4. If a number be divided into any two parts, the square of the number is equal to the sum of the squares of the two parts, and twice the product of those parts. THEOREM 5. If the difference of the cubes of any two numbers be divided by their difference, the quotient arising will be equal to the sum of the squares of the two numbers togeth- er with their product. THEOREM 6. When the sum and product of two numbers are given t» find the numbers * These Theorems should be committed to memory. QUESTIONS FOR EXERC1M . 9quare the sum of the numbers, and from that times the product, the square root of rence will give the difference of the numbers, then agreeably to Theorem 1, by having the sum and difference ,m ijmIv find the Dumb THEOREM 7. The sum, and sum of the squares of two numbers being given to find the numbers. Rule. — Subtract the sum of the squares of the num- bers, from the square of their sum, and half the remain- der will be their product. When the sum and product are given, we can find the numbers agreeably to Theorem (i. QUESTIONS FOR EXERCISE. 1. A wealthy man two daughters had, And both were very lair, To each he gave a tract of land, One round, the other square. At twenty shillings an acre just, h piece its value had; The shillings that did compass each For it exactly paid. If 'cross a shilling be an inch, As it is very near, Which was the greater fortune, She that had the round or square? Ans. Area of the square = 25090560 acres. Area of the circle = 19706125.82 10 acres. A. A bullet is dropped from the top of a building, and found to reach the ground in 2 seconds. Required it- height' Ans. 2 x 4. *= 8 and s x 8 = 64 feet l'>. Admitting I let fall a bullet from the top of a build- and found it reached the ground in 1 £ seconds. Re- quired its height? Ans. 36 ft. J. If a roll of butter weighs in one scale 2* lbs., and being changed into the other weighs 4 lbs.; what ii the true weight of the b> Ans. 3 lbs. QUESTIONS FOR EXERCISE. & A farmer hired two men, Reuben and Richard, for $30, (being the wages of both for 1 month,) to Reuben he s4 per month more than to Richard. How much did each get per month? Ans. Reuben got $17 and Richard $13. 4. There is a certain number which being divided by 11, the quotient resulting multiplied by 5, that product divided by 4, from the quotient subtract 75, to the remain- der add 40, and half the sum shall make 45. Required the number? Ans. 1100. 5. What is the discount of $400 for 2 years, 8 months, 24 days, at 6 per cent, per annum? Ans. $56.36. 6. What is the interest of $400 for 2 years, 8 months, 24 days, at 6 per cent.? Ans. $65.60. 7. A merchant in Baltimore received from New Or- leans a bill at 30 days sight; he allowed 1 "per cent, dis- count for present payment, and received $2530.44; what .sum was the bill drawn for; and what was the discount? Ans $ J + disct. $25.30. 8. Bunker Hill monument is 30 feet square at its base, and is to be 15 feet square at its top, its height is to be 220, from the bottom to the top through its centre, is a cylindrical avenue 1 5 feet in diameter at the bottom, and about 1 1 feet at the top. How many cubic feet will there be in the monument? Ans. 86068.518 -f ft. 9. Divide 40 into 4 such parts, that if to the first you add 4, from the second subtract 4, multiply the third by 4, and divide the fourth by 4; the sum, difference, pro- duct and quotient will be equal? Ans. l z z , s 5 *,} and *f 8 . 10. Find two numbers in a given ratio as 6 to 9, so that their sum and product may be equal? 2£ and 1$. 11. Two drovers, A and B, (from the state of Ohio) stopped at the "Three Tun Tavern," in the city of Bal- timore, each having a drove of cattle; A had 21 head, in his drove, and B 19; they agreed to join both together, and sell them for $27 per head; and, that A should receive Qtr III. 277 $3 a head more than B; how much did each get for his Ans. A got #596.92 j; B $483.07J. 1 -J. What number u it to which J of $ of $ be added, the sum will be 1? Ans. §£. ] 3. Wi. it number is it whose ^, a and \ added togeth* make 78? Ans. 72. 1 \. To find a number, which being multiplied by 3, subtracting 5 from the product and the remainder divided by 2, if the number sought be added to the quotient, the sum will be 40? Ans. 17. 15. What number is it which being added to 4, and also multiplied by 4, the product shall be treble the sum? Ans. 12. 16. To find a humber, to which if 11 be added and 7 subtracted from the same number, the sum of the addition will be double the remainder? Ans. 25. 17. To find a number which being added to itself and the sum being multiplied by the same, and the same num* ber still subtracted from the product; and lastly, the re* mainder divided by the same, the quotient may be 13? Ans. 7. 18. To find three numbers such, that the sum of the first and second shall be 15, the sum of the first and third 'id the sum of the second and third 17? Ans. 7, 8 and 9. 19. Three persons, A, B and "C, owe certain sums of money, so that A and B owe $210, B and C 290, and C and A 400; what did each of them owe? • Ans. A 160, B 50, and C 240. 20. A man having a certain number of dollars in his hand, being asked how many he had, replied if the num- ber be divided by 5 and 19 added to the quotient, I shall then have $23; how many dollars had he? Ans. 20. 21. A person said he had 20 children, and that it hap- pened there was a year and a half between each of their -age*, his eldest was born when he was 24 years old, and 24 278 QUESTIONS FOR EXERCISE. the age of the youngest is now 21; what was the fathere f age? Ans. 73$ years. 22. When I sold cloth at $7 per yard, I gained 56J cents; what would I gain by selling 3 pieces, which cost ine $4.00? Ans. $32. 14*. 23. A and B dissolve partnership, and equally divide their gain; A's share which was $332.50 lay for 21 months; B's for 9 months only; the adventure of B is required? Ans. $775.83 J. 24. I bought 25 yards of lace at 30 cents for the iirst yard, and for. the last yard 96; the price of each yard in- creasing in arithmetical progression; what did the whole amount to? Ans. $15.75. 25. A case of dry goods amounting to .£230 5*. ster- ling is sold in Philadelphia, at 20 per cent, advance; what is the amount in American currency? Ans. $1228. 26. What is the difference between the interest of $1000 at 6 per cent, for 8 years, and the discount of the same sum at the same rate, and for the same time? Ans. the interest exceeds the discount by $155.67$. 27. A gentleman sells a piece of cloth at $56.25, and thereby loses 1\ per cent.; what was the first cost? Ans. $60.47. 28. A has 28 cwt of cotton which cost him $264-; how must he rate it per lb., to D, so that by taking his note payable at 9 months he may clear $30, allowing in- terest at 6 per cent, per annum? Ans. 9|c. 29. In a series of proportional numbers the first is 5, the second is 8, the product of the second and third is 78.4, what is the difference of the first and fourth? Ans. 10.68. 30. If 3481 soldiers are to be placed in square file, how many are to be set in each rank? Ans. 59. 31. A can do a piece of work in 10 days, B in 13, set them both at it, in what time will it be finished? Ans. 5||. QUESTIONS F©R EXERCISE. 279 A stationer sold quills at $10 per thousand, by which he gained 2 of the money, but growing scarce raised them to $12,50 per thousand, what is the gain per cent by the latter price? Ans. 100 per ct. B& Richard Cotter, of Baltimore, remits to William Denman, of New York, a bill of exchange on London, the avails of which he wishes to be invested in goods on his account. Denman disposes of the bill at 7£ percent. advance, and received $9675.00, having reserved for himself i per cent, on the sale of the bill, and 2 per cent, for commission, what will remain for investment, and how much was the bill drawn for? Ans. for investment $9457.311. The bill was drawn for £2025 sterling. SOUM) Sound if not interrupted, will move at the rate of about 1150 feet in a second of time, 34. I have seen the flash of a cannon fired from Fort _Mc Henry, and heard the report 47 seconds afterwards, what distance was the fort from where I stood? Ans. 54050 feet. 35. Bought a quantity of cloth for $750, \ of which I found to be inferior, which I had to sell at $1.25 per yard, and by this I lost $100, what must I sell the rest at per yard, that I shall lose nothing by the whole? Ans. $3.15^f. 36. A circular fish pond is to be dug in a garden that shall take up just half an acre, what must be the radiut of the circle? Ans. 27.75 yds. 37. Bought a horse which was worth 30 per cent, more than I gave for him, but having been injured I sold him for 25 per cent, less than what he cost, and thereby lost $55 on his real value, what was received for the horse? Ans. $75. 38. A son having asked his father's age the father thus replied; "your age is 8 years, to which if five-eights of QUESTIONS FOR EXERCISE. both our ages be added, the sum will be my age," what was the father's age? Ans. 34 years, 8 months. 39. A merchant sold goods to a certain amount on a commission of 4 per cent, and having remitted the net proceeds to his correspondent, he received i per cent, for prompt payment, which amounted to $15.60, what was the amount of his commission? Ans. $260.00. 40. How many thousand brick would be required to build the walls of a house 40 feet long, 30 feet wide, and 20 feet high, admitting the walls to be a foot thick, and that each brick was 8 inches long, 4 inches wide, and 2 inches thick? Ans. 73440. 41. Purchased merchandise to the amount of $2000, viz: $400 at 3 months, $800 at 6 months, and $800 at 9 months. Required the average time of payment. Ans. 6} months. Bought several parcels of goods at different times and on various crec 1841. Al.iv 1 1, a bill am'ting to $75 at 1 mo. crd. due June 10. « 18, " « 64 « 2 « 25, * " 96 « 3 June 6, « " 104 « 4 < " 20, " « 144 " 5 Required the average time of payment? Ans. July 14th, at 9 o'clock in the morning. 43. The ball on the top of St. Paul's Church is 6 feet in diameter, what did the gilding of it cost at 3£d. per square inch? Ans. j£237 10s. Id. 44. There is a conical glass, 6 inches high, 5 inches wide at the top, and is ± part filled with water, what must be the diameter of a ball, let fall into the water> that shall be immersed by it? Ans. 2.445 -f in. 45. How much larger is a cube, that will contain a globe of 20 inches in diameter, than a cube inscribed within such globe? Ans. lar. 8000 in. smaller 1539 iiu <t a " July 17. u u « Aug. 23. a a " Oct. 4. U « " Nov. 17. SUBSCRIBERS' NAMES. CLERGYMEN. Rev. G. Raymond, D. D. Presi- dent St. Mary's College " E. A. Knight, V. Pres't • L. R. Deluol Hugh Griffin " F. Lhomme P. Courtney B. Daraphoux, St. Jo- seph's Church " John B. Gildea, St. James' " P. S. Schreiber, Cathedral ■• (has. J. White, seph Prost, St. John's fr. Kerney, St. Patrick's Fell's Point •• .l.iiiit'* Dolan, " " • David Steele, Exeter st. " John F. Hey, Fayette st. <; John Smith, Pastor of Sea- men's Bethel, F. P. - unuel Kepler, Wilk st. » M'rard Morgan, Car. st. .lohn A.Gere, Aisquithst. John Valiant, Wiliest. lohn Watts, Lexington st. " John Russel, (German ngelical) " Henry Scheib (German Lutheran /ions' Church) TEACHERS. John Neely, A. B. Hanover st. John Magee, Esq. Fred'k st. Hugh Bannan " Edmund Smith " David Rinc, E. B. F. Institute Patrick H. M'Sorley N. Spelman Robert S. Wrixon J. J. Reekers, prof, of M. & L. a alker, prof, of Math. 1 N MJilton,Esq.20Eutawst. Frederick Foy James E. Sea'rly Henry Haunt James McCrodan Shepard C. Keith James McCarrick James Mclntire, A. M. James Harshaw, A. M. John Galvin, M. C. T. George Sharp Richard Cotter, Esq. Professor of Mathematics Z. George, prof, of Languages John H. Shea Henry McCann Edward Hazen, Author S. H. Matthews N. C. Brooks, A. M. Principal Hiiih School in Baltimore. P. Kelly, professor St. Mary's Seminary LAWYERS. Hon. R. B. Taney, C. J. U. S. Hon. Sol. Hillen, M. C. James M. Buchanan Wm. P. Preston Wm. H. Watson Wm. C. Collins John C. Legrand Chas. H. Pitts Geo. R. Richardson T. Parkin Scott Wm. Gwynn Chas. Z. Lucas Samuel Stump, jr. William F. Giles John Scott PHYSICIANS. H. A. Inloes. John C. S. Monkur John F. Monmonier Nathaniel Potter P. S. Kinneman Henry Morris Charles Maguire Samuel G. Baker Nathan R. Smith Richard Wiltnot Hall Levi D. Bodder Jas. Mclntire William E. A. Aikin Daniel H. Lawrence D. P. Chatard John H. O'Donovan. Andrew A. Lynch G. Schwarz SUBSCRIBERS NAMES. II W. Waters (Maria Grove) T. L. Murphy (Bond st. F. P.) MAGISTRATES. Jeremiah Storm Ninon Kemp W". king Wm. Pechin Win. B. Jones John Burnham John Wright M. Bod) I Walsh James Blair Peregrine Gorsuch J as. Barnard, jr. Jas. P. Heath Henry R. Lauderman A. H". Pennington Chas. Keenan <ier John Walter Eugene O'Connor NOTAR1 Robert M W William Kichelberger \ 1KRS. Win. ( Richard J. Matchett J no. Murphy, 146 Market st. Sherwood &. Co. James Young A. H. Simmons, Ed. "Sun'* BOOK SELLERS. David Owen, 24 N. Gay st. Geo. P. Knotts, Market st. F. P. Knight & Colburn MERCHANTS. Meredith, Spencer & Co., 246 Baltimore st. P. B. Curtiss, Han. & Lomb. J.W. Randolph, Market Wm. Jackson, Thames st Wm. Wilson &. Sons, 105 Mar- ket st. T. J. Ahern.Liberty&Balt.sts. Elijah Stansbury Francis Hyde & Sons A. Penniman, North st. John O'Brien Daniel Coonan Benjamin Thomas James Dickson B J & E J Saunders James Hooper, Ball. st. Ceo. Waters &. Son, F. P. Wm. Gibson, Lewis Auodoun ,iel Dorsey M. Stillinger Isaac Hartman Patrick Mullin Wm. G. Bennett John Bennett John Thompson Samuel Wilson 1120 D. Taylor Wm. A. Sloan Thos. J. Warrington Edwd. W. Bennett Turner, Wheelwright & Mudge John J. Smith, accountant O'Connor & Co., North st. M. Blundell, 7 1 ird st. Wm. Wilson, 16 Franklin st. i: McFaul &, Co., Paca st. MERCH \M > l*d GROCERS. Thos. R. Simmons Alfred Wiles Washington Brown G. M. Hutton, 24 Baltimore st. John T. Laws, accountant Robert Barry, I Lemuel A. Gordon Robert Clark John I. Gross, auctioneer George Presstman Thos. E. Reardon Edwd. W. Bennett Thomas Walsh, 18 S. Ga\ Alexander Ager, Second st. John T. Burns, accountant Thomas P. Jenkins, Gay st. David C. Springer, Pratt st George H. Keys, Gough st. John T. Spencer James A. Morris John Drum Justus Bruehl Edward P. Breonaa John Bates B. McDonough John Matthews. .-,9 S. Charles st. Jeremiah O'Brien O. F. Pettit James M'Court Wm. H. Fowler Wm. Warden burg Samuel P. Curry Mich'l McGrath.Eutaw &Mad, Frederick Eversmau SUBSCRIBERS .TAMES. Thoo. Morxlock Willi* John <;. Tebebnan William Solomon John W. Croney k Ha-. in .• tn Abbott K. Stansbury B. Curry Marti John Hall John S turhoff -es WiesenfVl.l Thomas Leib, jr. Samuel Swobe Philip 1YI1 Emanuel Weinman Joseph Zinkand I Shuiiian Robt. Thompson James Diggs A nthony Kenan, Union Hotel Edwd. Boyle, 9th Ward House P. Adams, Hotel, 2 Light st. James Brow -n. Washington hotel Michael McCaraher.'Sarn urn's' apotiikc uur.s. Thos. J. Pitt James P. Williamson Joseph B. Stansburv Ml NlClPAL OFFICElS. Hon, S. C. Leakin Hon. Samuel Brady, Mayor John Mitchell, H. C. H. G. Miller M. Benzinger, Esq. Jacob Small, Esq. -bury CITY COUNCIL. John E. Stansbury Joseph Rai: James Fields « H'Y OFFICERS. Jeremiah Frazier Albert C. Hall William A. Swill I lor SE I C Ml ilJSi and OFFI- - SERVICE. Win. S. Coiinland, E. Pratt st. Joshua Fernll Michael McDonald John M'Gowan, U. S. i - Frazier u u .Stafford John Hesse, U. S. R. S. Chas. B. U.-a.Hort, ■• Robert Hcnthall Gay Moore, U. S. R. S. Wm. Wall, U.S. A. H. Prince H. Davey Wm. Thompson James Gibson Jo»eph Ennis Wm. McLennan W.>. Wedge James Gunhv Wm. C. Lewis R. Robinson Thos. T. Wingale Edwin Bailey Henry H. Willis Lieut. Sands, U. S. R. S. " Beverly Diggs " " A. Burrough J. J. B. Walbach, U. S. N. J. M. Gardner, " " James Kelly, U. S. A. COMMODORE DAN IF, Li- C. H. Cooksey James Beard J, B, Corner Samuel Keene Samuel Wall John H. Fry John S. Oliver James Giieves John Dutton Thomas Kelly Edward Gouid Wm. B. Higgina John Myers Jacob W. Hugg George Little Walter Oram John Martin PILOTS. Hector Jackson Robert M. Bear James Cheveral Thos. G. L. Clark James A. Milbouru J. P. Hancy William Chase Thomas Lewi* John Cost subscribers' JTAMfcS, HATTEKS. Chas. Townsend Levi Lowe P. Mil: J. Madison Anderson J. S. Horton Geo* W. Tavlor> North st. No. 8 PAINTERS. Richard R. Bishop, Thames st. M. O'Laughlin, cor. Holliday and Fayette st. JohnDobson, Wolf st. F. P. John Welch, Ann st. Duvall &. Maccubbin,Fayette st. Win. Johnson, 73 Lombard st* PLUMBERS & COPPER- BMITH8. , H. Hall H.nrv Boa, Thames st. James Kelly, 53 South st. John M. Bruce, Light st. Benj. F. Barkman, Thames st. Edward G. Dorry, Thames st. Frederick Daiger, Pratt, st. Frmacil Kines, Light st. COOPERS. Nicholas Gill, near Eutaw bouse Wm. S. Espy &. Co., 27 Ann st. Jos. Golibart, sr., Alice Ann st. Jos. M'Dermott, Philpot st. Benj. Abbott, President st. Alex. Logan, Lerew's alley Thos. Woollen, Commerce st. John K. Carroll, Martin Corban 4< Jos. Golibart, jr., Ann st. Henry Hush, Howard st. Robert Wilson, Fayette st. Reuben Hicks, German st. J. W. Allen, German near Paca George H. Watts, 272 Market st. Henry Cook. Lee st. Wm. Patterson, " Richard Pierce James Pouch Wm. Douglass* Green st. COACH MAKERS. James McCurley, Liberty st. Wm. Selvage, Fayette st. James Stewart, Calvert st Robert Leckie, Philad. Road Wm. McCann, Frederick st. Thomas C. Dunlevy* Gay st. CUSTOM HOUSE OFFK I Win. Frick, collector J . K. Law, Esq., dep. " Lyon, |r .1. W. Wilmer,7Bowly'swharf Henry \\ Charles Soran Wm. V. Jenkins ge W. Uurke.Founiainrow Richard Mackubbin Thomas Kyle, F. P. 1 ). \V. Hudson, 10 Fountain row .Mm .Niints 72 High it Robert Aitken, 191 E. Pratt st. James Jenkins, Lovely Lane John Lowry> Mechanics' I W.Ik st. property A<;r\ Jas. Tracey, St. Paul Geo. Gelbach, broker, N.W. cor. Baltimore and Frederick st. Abm. Cuyk, 42 Fell st., F. P Chas. Dieselhorst, 24 Thames st. watchmaki:k>. Get). A Syberts, Bond st. John Somer Riefner, " Aaron I " Jas. V. Brown, " A. Ostendorff, Camden st. Everhard Krimmer Robert Holloway, High st. CABINET MAKERS, rii S. Tarr,4Gay st. Wesley B. Tarr, 32 " R. A. L. Bevans, E. Balto. st. John Hughes, Bond st. John Garrett, Alice Ann st. Jas. P. Weakley, Pitt st. Jas. Byrne, Fred. & Lom. st. Jacob Rulf, Caroline st. George Hoffman, Park st. Wm. King, 72 Charles st. CARPENTERS. John Roy, Hanover st. Thos. C. Monmonier, F. P. Wm. Campbell, Washington St. Geo. A. Heuisler, Madison st. W. E. Beale,Silver & Spring st. E. W. Robinson,Marketst.F.P. John R. Budd, Alice Ann st. John M'Ardle, York avenue Mich. Flaherty, Washington st. H. W. S. Evans, cor. Gough k Eden st. I YU 35827 3A102 Educ . Lib. , % HI I I i